1^- 


N     ? 


ELEMENTS 


OF 


PLANE    AND    SPHERICAL 

TRIGONOMETRY, 

WITH   THEIK   APPLICATIONS   TO 

MENSURATION,  SURVEYING,  AND 
NAVIGATION. 


BY    ELIAS    LOOMTS,    LL.D., 

PllOFESSOr.    OF   NATtTEAL   PUILOSOPHY   AND    A8TBONOMY    IN   YALE    COLLEOK,  AND    AUTHOR    OP 
A  "^OOUESE   OF   MATHEMATICS." 


FOURTfcElCTH*  EDITJOlC 


NEW    YORK: 
HARPER    &    BROTHERS,    PUBLISHERS, 

FKANKLIN    SQUARE. 

1862. 


Entered,  according  to  Act  of  Congress,  in  the  year  one  thousand 
eight  hundred  and  fifty-eight,  by 

Harper  &  Brothers, 

in  the  Clerk's  Office  of  the  District  Court  of  the  Southern  District 
of  New  York. 


T  R  E  F  A  C  E. 


TuE  following  treatise  constitutes  the  third  volume  of  a 
course  of  Mathematics  designed  for  colleges  and  high  schools, 
and  is  prepared  upon  substantially  the  same  model  as  the  works 
on  Algebra  and  G-eometry.  It  does  not  profess  to  embody 
every  thing  which  is  known  on  the  subject  of  Trigonometry, 
but  it  contains  those  principles  which  are  most  important  on 
account  of  their  applications,  or  their  connection  with  other 
parts  of  a  course  of  mathematical  study.  The  aim  has  been 
to  render  every  principle  intelligible,  not  by  the  repetition  of 
superfluous  words,  but  by  the  use  of  precise  and  appropriate 
language.  Whenever  it  could  conveniently  be  done,  the  most 
important  principles  have  been  reduced  to  the  form  of  theorems 
or  rules,  which  are  distinguished  by  the  use  of  italic  letters, 
and  are  designed  to  be  committed  to  memory.  The  most  im- 
portant instruments  used  in  Surveying  are  fully  described,  and 
are  illustrated  by  drawings. 

The  computations  are  all  made  by  the  aid  of  natural  num- 
bers, or  with  logarithms  to  six  places ;  and  by  means  of  the 
accompanying  tables,  such  computations  can  be  performed 
with  great  facility  and  precision  This  volume,  having  been 
used  by  several  successive  classes,  has  been  subjected  to  the 
severest  scrutiny,  and  the  present  edition  embodies  all  the  al- 
terations which  have  been  suggested  by.  experience  in  the  re- 
citation roona. 


33fJ^S9 


CONTENTS 


BOOK  I. 

THE  NATURE  AND  PROPERTIES  OF  LOGARITHMS. 

Nature  of  Logarithms 7 

Description  of  the  Table  of  Logarithms 9 

Multiplication  by  Logarithms 14 

Division  by  Logarithms li 

Involution  by  Logarithms - 17 

Evolution  by  Logarithms *...  17 

Proportion  by  Logarithms 18 

BOOK  II. 

PLANE  TRIGONOMETRY. 

iSines,  Tangents,  Secants,  &c.,  defined 20 

Explanation  of  the  Trigonometrical  Tables 23 

To  find  Sines  and  Tangents  of  small  Arcs 29 

Solutions  of  Right-angled  Triangles 32 

Solutions  of  Oblique-angled  Triangles 36 

Instraments  used  in  Drawing _ , 42 

Geometrical  Construction  of  Triangles 46 

Values  of  the  Sines,  Cosines,  &c.,  of  certain  Angles 48 

Trigonometrical  Formulao 52 

Computation  of  a  Table  of  Sines,  Cosines,  &c 57 

BOOK  III. 

MENSURATION  OF  SURFACES  AND  SOLIDS. 

Areas  of  Figures  bounded  by  Right  Lines 59 

Area  of  a  Regular  Polygon 64 

Quadrature  of  the  Circle  and  its  Parts 66 

Mensuration  of  Solids 71 

Rail-way  Excavations  or  Embankments 77 

Regular  Polyedrons 81 

The  three  Round  Bodies 84 

Area  of  a  Spherical  Triangle 88 

BOOK  IV. 

SURVEYING. 

Definitions M 


vi  Contents. 

Pag. 

Explanation  of  the  Vernier 94 

Description  of  the  Theodolite 95 

Heights  and  Distances 97 

The  Determination  of  Areas 103 

Plotting  a  Survey 104 

The  Traverse  Table 106 

To  find  the  Area  of  a  Field 109 

Trigonometrical  Surveys 114 

Variation  of  the  Needle 117 

Leveling  .* 119 

Topographical  Maps 123 

Setting  out  Rail-way  Curves 127 

Surveying  Harbors 130 

The  Plane  Table...... 132 

To  determine  the  Depth  of  Water.... , 133 

BOOK  V. 

NAVIGATION. 

Definitions,  &c 135 

Plane  Sailing 138 

Traverse  Sailing 141 

Parallel  Sailing 144 

Middle  Latitude  Sailing 14G 

Mercator's  Sailing 149 

Nautical  Charts 153 

BOOK  VI. 

SPHERIGAL  TRIGONOMETRY. 

Right-angled  Spherical  Triangles 15& 

Napier's  Rule  of  the  Circular  Parts 158 

Examples  of  Right-angled  Triangles IGO 

Oblique-angled  Spherical  Triangles 1G3 

Examples  of  Oblique-angled  Triangles 1G5 

Trigonometrical  Formulse 171 

Sailing  on  an  Arc  of  a  Great  Circle 17fi 


TRIGONOMETEY, 


BOOK  I. 

THE  NATURE  AND  PROPERTIES  OF  LOGARITHMS. 

Article  1.  Logarithms  are  numbers  designed  to  diminish 
the  labor  of  Multiplication  and  Division,  by  substituting  in  their 
stead  Addition  and  Subtraction.  All  numbers  are  regarded  as 
powers  of  same  one  number,  which  is  called  the  base  of  the 
system ;  and  the  exponent  of  that  power  of  the  base  which  is 
equal  to  a  given  number,  is  called  the  logarithm  of  that  number. 

The  base  of  the  common  system  of  logarithms  (called,  from 
their  inventor,  Briggs'  logarithms)  is  the  number  10.  Hence 
all  numbers  are  to  be  regarded  as  powers  of  10.  Thus,  since 
10°= 1,  0  is  the  logarithm  of  1  in  Briggs'  system; 

10^=10,         1      "  "  10  "  " 

10^=100,       2      "  "  100  "  " 

10^=1000,     3      "  ''  1000  "  " 

10*=10000,  4      "  "  10,000  "  " 

&c.,  &c.,  &c. ; 

whence  it  appears  that,  in  Briggs'  system,  the  logarithm  ol 
every  number  between  1  and  10  is  some  number  between  0 
and  1,  i.  e.,  is  a  proper  fraction.  The  logarithm  of  every  num- 
ber between  10  and  100  is  some  number  between  1  and  2,  i.  e., 
is  1  plus  a  fraction.  The  logarithm  of  every  number  between 
100  and  1000  is  some  number  between  2  and  3,  i.  e.,  is  2  plus 
a  fraction,  and  so  on. 

(2.)  The  preceding  principles  may  be  extended  to  fraction«i 
by  means  of  negative  exponents.     Thus,,  since 


10-^=0.1, 

—  1  is  the 

logarithm 

of  0.1       in 

Briggs'  system  t- 

10-^=0.01, 

-2     M 
-3    A 

li 

0.01 

U                    li 

10-' =0.001, 

u 

0.001 

ti                         4( 

10-^=0.0001 

-4     " 

»( 

0.0001 

((                         if 

&c., 

&c., 
A 

&0. 

8  Trigonometry. 

Hence  it  appears  that  the  logarithm  of  every  numLer  between 
1  and  0.1  is  some  number  between  0  and  —1,  or  may  be  rep 
resented  by  —1  plus  a  fraction;  the  logarithm  of  every  num- 
ber between  0.1  and  .01  is  some  number  between  —1  and  —2, 
or  may  be  represented  by  —2  plus  a  fraction;  the  logarithm 
of  every  number  between  .01  and  .001  is  some  number  be- 
tween —2  and  —3,  or  is  equal  to  —3  plus  a  fraction,  and 
so  on. 

The  logarithms  of  most  numbers,  therefore,  consist  of  an  in- 
teger and  a  fraction.     The  integral  part  is  called  the  char  at 
teristic^  and  may  be  known  from  the  following 

Rule. 

The  characteristic  of  the  logarithm  of  any  number  greate? 
than  unity,  is  one  less  than  the  number  of  integral  figures  in 
the  given  number. 

Thus  the  logarithm  of  297  is  2  plus  a  fraction  ;  that  is,  the 
characteristic  of  the  logarithm  of  297  is  2,  which  is  one  less 
than  the  number  of  integral  figures.  The  characteristic  of  the 
logarithm  of  5673.29  is  3  ;  that  of  73254.1  is  4,  &c. 

The  characteristic  of  the  logarithm  of  a  decimal  fraction 
is  a  negative  number^  and  is  equal  to  the  number  of  places  by 
which  its  first  significant  figure  is  removed  from  the  place 
of  units. 

Thus  the  logarithm  of  .0046  is  —3  plus  a  fraction ;  that  is, 
the  characteristic  of  the  logarithm  is  —3,  the  first  significant 
figure,  4,  being  removed  three  places  from  units. 

(3.)  Since  powers  of  the  same  quantity  are  multiplied  by 
adding  their  exponents  (Alg.,  Art.  50), 

The  logarithm  of  the  product  of  two  or  more  factors  is 
equal  to  the  sum  of  the  logarithms  of  those  factors. 

Hence  we  see  that  if  it  is  required  to  multiply  two  or  more 
numbers  by  each  other,  we  have  only  to  add  their  logarithms : 
the  sum  will  be  the  logarithm  of  their  product.  "We  then  look 
in  the  table  for  the  number  answering  to  that  logarithm,  in 
order  to  obtain  the  required  product. 

Also,  since  powers  of  the  same  quantity  are  divided  by  sub- 
tracting their  exponents  (Alg.,  Art.  66), 

The  logarithm  of  the  quotient  of  one  number  divided  by  an- 


Logarithms.  9 

Jther,  is  equal  to  the  difference  of  the  logarithms  of  those 
numbers.  . 

Hence  we  see  that  if  we  wish  to  divid,e  one  number  by  an- 
other, we  have  only  to  subtract  the  logarithm  of  the  divisor 
from  that  of  the  dividend  ;  the  difference  will  be  the  logarithm 
of  their  quotient. 

(4.)  Since,  in  Briggs'  system,  the  logarithm  of  10  is  1,  if 
any  number  be  multiplied  or  divided  by  10,  its  logarithm  will 
be  increased  or  diminished  by  1 ;  and  as  this  is  an  integer,  it 
will  only  change  the  characteristic  of  the  logarithm,  without 
affecting  the  decimal  part.     Hence 

The  decimal  part  of  the  logarithm  of  any  number  is  the 
same  as  that  of  the  number  multiplied  or  divided  by  10,  100. 
1000,  &c. 

Thus,  the  logarithm  of  65430  is  4.815777 

^'  6543  is  3.815777 

"  654.3  is  2.815777 

"  "  65.43  is  1.815777 

"  ''  6.543        is  0.815777 

"  "  .6543      is  1.815777 

^  "  "  .06543    is  2.815777 

"  "  .006543  is  3.815777. 

The  minus  sign  is  here  placed  over  the  characteristic,  to 
show  that  that  alone  is  negative,  while  the  decimal  part  of  the 
logarithm  is  positive. 

Table  of  Logarithms. 

(5.)  A  table  of  logarithms  usually  contains  the  logarithms 
of  the  entire  series  of  natural  numbers  from  1  up  to  10,000, 
and  the  larger  tables  extend  to  100,000  or  more.  In  the  smaller 
tables  the  logarithms  are  usually  given  to  five  or  six  decimal 
places  ;  the  larger  tables  extend  to  seven,  and  sometimes  eight 
or  more  places. 

In  the  accompanying  table,  the  logarithms  of  the  first  100 
numbers  are  given  with  their  characteristics  ;  but,  for  all  other 
numbers,  the  decimal  part  only  of  the  logarithm  is  given,  while 
the  characteristic  is  left  to  be  supplied,  according  to  the  rule 
in  Alt.  2. 


10  Trigonometry. 

(6.)  To  find  the  Logarithm  of  any  Number  between  1  and  100. 
Look  on  the  first  page  of  the  accompanying  tahle,  along  the 
column  of  numbers  under  N.,  for  the  given  number,  and  against 
it,  in  the  next  column,  will  be  found  the  logarithm  with  its 
characteristic.     Thus, 

opposite  13  is  1.113943,  which  is  the  logarithm  of  13 ; 
"       Q>^  is  1.812913,  "  "  65. 

To  find  the  Logarithm  of  any  Number  consisting  of  three 

Figures. 

Look  on  one  of  the  pages  of  the  table  from  2  to  20,  along 
%e  left-hand  column,  marked  N.,  for  the  given  number,  and 
against  it,  in  the  column  headed  0,  will  be  found  the  decimal 
part  of  its  logarithm.  To  this  the  characteristic  must  be  pre- 
fixed, according  to  the  rule  in  Art.  2.  Thus 
the  logarithm  of  347  will  be  found,  from  page  8,  2.540329; 

"  "  871  "  ''       18,  2.940018. 

As  the  first  two  figures  of  the  decimal  are  the  same  for  sev 
eral  successive  numbers  in  the  table,  they  are  not  repeated  for 
each  logarithm  separately,  but  are  left  to  be  supplied.  Thus 
the  decimal  part  of  the  logarithm  of  339  is  .530200.  The  first 
two  figures  of  the  decimal  remain  the  same  up  to  347 ;  they 
are  therefore  omitted  in  the  table,  and  are  to  be  supplied. 

To  find  the  Logarithm  of  any  Number  consisting  of  foin 

Figures. 

Find  the  three  left-hand  figures  in  the  column  marked  N., 
as  before,  and  the  fourth  figure  at  the  head  of  one  of  the  other 
columns.  Opposite  to  the  first  three  figures,  and  in  the  col- 
umn under  the  fourth  figure,  will  be  found  four  figures  of  the 
logarithm,  to  which  two  figures  from  the  column  headed  0  are 
to  be  prefixed,  as  in  the  former  case.  The  characteristic  must 
be  supplied  according  to  Art.  2.     Thus 

the  logarithm  of  3456  is  3.538574 ; 
"  "  8765  is  3.942752. 

In  several  of  the  columns  headed  1,  2,  3,  &c.,  small  dots  are 
found  in  the  place  of  figures.  This  is  to  show  that  the  two 
figures  which  are  to  be  prefixed  from  the  first  column  have 
changed,  and  they  are  to  be  taken  from  the  horizontal  line  di- 


Logarithms.  IJ 

rectly  below.     The  place  of  the  dots  is  to  be  supplied  with  ci- 
phers.    Thus 

the  logarithm  of  2045  is  3.310693  ; 

"  "  9777  is  3.990206. 

The  two  leading  figures  from  the  column  0  must  also  be 

taken  f -om  the  horizontal  line  below,  if  any  dots  have  been 

passed  over  on  the  same  horizontal  line.     Thus 

the  logarithm  of  1628  is  3.211654. 

To  find  the  logarithm  of  any  Number  containing  more  than 
four  Figures. 
(7.)  By  inspecting  the  table,  we  shall  find  that,  within  cer- 
tain limits,  the  differences  of  the  logarithms  are  nearly  propor- 
tional to  the  differences  of  their  corresponding  numbers.    Thus 
the  logarithm  of  7250  is  3.860338 
7251  is  3.860398 
''  "  7252  is  3.860458 

«'  "  7253  is  3.860518. 

Here  the  difference  between  the  successive  logarithms,  called 
the  tabular  difference,  is  constantly  60,'  corresponding  to  a  dif- 
ference of  unity  in  the  natural  numbers.  If,  then,  we  sup- 
pose the  logarithms  to  be  proportional  to  their  corresponding 
numbers  (as  they  are  nearly),  a  difference  of  0.1  in  the  num- 
bers should  correspond  to  a  difference  of  6  in  the  logarithms ; 
a  difference  of  0.2  in  the  numbers  should  correspond  to  a  dif- 
ference of  12  in  the  logarithms,  &c.     Hence 

the  logarithm  of  7250.1  must  be  3.860344 ; 
'    "  "  7250.2         "      3.860350; 

"       ,     "  7250.3        "      3.860356. 

In  order  to  facilitate  the  computation,  the  tabular  difference 
is  inserted  on  page  16  in  the  column  headed  D.,  and  the  pro- 
portional part  for  the  fifth  figure  of  the  natural  number  is  given 
at  the  bottom  of  the  page.  Thus,  when  the  tabular  difference 
is  60,  the  corrections  for  .1,  .2,  .3,  &c.,  are  seen  to  be  6,  12, 
18,  &c. 

If  the  given  number.was  72501,  the  characteristic  of  its  log- 
arithm would  be  4,  but  the  decimal  part  would  be  the  same  a« 
for  7250.1. 

If  it  were  required  to  find  the  correction  for  a  sixth  figure 


12  Trigonometry. 

in  the  natural  number,  it  is  readily  oLtained  from  the  Propoi- 
tional  Parts  in  the  table.  The  correction  for  a  figure  in  the 
sixth  place  must  he  one  tenth  of  the  correction  for  the  same 
figure  if  it  stood  in  the  fifth  place.  Thus,  if  the  correction  for 
.5  is  30,  the  correction  for  .05  is  obviously  3. 

As  the  differences  change  rapidly  in  the  first  part  of  the  ta- 
ble, it  was  found  inconvenient  to  give  the  proportional  parts 
for  each  tabular  difference ;  accordingly,  for  the  first  seven 
pages,  they  are  only  given  for  the  even  differences,  but  the  pro- 
portional parts  for  the  odd  differences  will  be  readily  found  bv 
inspection. 

Required  the  logarithm  of  452789. 

The  logarithm  of  452700  is  5.655810. 

The  tabular  difference  is  96. 

Accordingly,  the  correction  for  the  fifth  figure,  8,  is  77,  and 
for  the  sixth  figure,  9,  is  8.6,  or  9  nearly.  Adding  these  cor- 
rections to  the  number  before  found,  we  obtain  5.655896. 

The  preceding  logarithms  do  not  pretend  to  be  perfectly 
exact,  but  only  the  nearest  numbers  limited  to  six  decimal 
places.  Accordingly,  when  the  fraction  which  is  omitted  ex- 
ceeds half  a  unit  in  the  sixth  decimal  place,  the  last  figure 
must  be  increased  by  unity. 

Required  the  logarithm  of  8765432. 

The  logarithm  of  8765000  is  6.942752 

Correction  for  the  fifth  figure,  4,  20 

"  "       sixth  figure,  3,  1.5 

"  "       seventh  figure,  2,  0.1 

Therefore  the  logarithm  of  8765432  is  6.942774. 

Required  the  logarithm  of  234567. 

The  logarithm  of  234500  is  5.370143 

Correction  for  the  fifth  figure,  6,  111 

"  '<•       sixth  figure,  7,  13 

Therefore  the  logarithm  of  234567  is       5.370267. 

To  find  the  Logarithm  of  a  Decimal  Fraction 

(8.)  According  to  Art.  4,  the  decimal  part  of  the  logarithm 
of  any  number  is  the  same  as  that  of  the  number  multiplied 
or  divided  by  10,  100,  1000,  &c.     Hence,  for  a  decimal  frao- 


Logarithms.  13 

tion,  we  find  the  logarithm  as  if  the  figures  were  integers,  and 
prefix  the  characteristic  according  to  the  rule  of  Art.  2. 


Examples. 

The  logarithm  of  345.6 

is  2.538574 

"            "             87.65 

is  1.942752 

"            "              2.345 

is  0.370143 

"             "                 .1234 

is  1.091315 

"  "  .005678  is  3.754195. 

To  find  the  Logarithm  of  a  Vulgar  Fraction. 
(9.)  "We  may  reduce  the  vulgar  fraction  to  a  decimal,  and 
fiwd  its  logarithm  hy  the  preceding  article  ;  or,  since  the  value 
of  a  fraction  is  equal  to  the  quotient  of  the  numerator  divided 
by  the  denominator,  we  may,  according  to  Art.  3,  subtract  the 
logarithm  of  the  denominator  from  that  of  the  numerator ; 
the  difference  will  he  the  logarithm  of  the  fraction. 
Ex.  1.  Find  the  logarithm  of  y%,  or  0.1875. 

From  the  logarithm  of  3,  0.477121, 

Take  the  logarithm  of  16,  1.204120.  . 

Leaves  the  logarithm  of  /_,  or  .1875,  1.273001.  .  i 

Ex.  2.  The  logarithm  of  /^  is  2.861697.  ' 

Ex.  3.  The  logarithm  of  iff  is  1.147401. 

To  find  the  Natural  Number  corresponding  to  any  Logarithm . 

(10.)  Look  in  the  table,  in  the  column  headed  0,  for  the  first 
two  figures  of  the  logarithm,  neglecting  the  characteristic  ;  the 
other  four  figures  are  to  be  looked  for  in  the  same  column,  o) 
in  one  of  the  nine  following  columns  ;  and  if  they  are  exactly 
found,  the  first  three  figures  of  the  corresponding  number  will 
be  found*  opposite  to  them  in  the  column  headed  N.,  and  the 
fourth  figure  will  be  found  at  the  top  of  the  page.  This  number 
must  be  made  to  correspond  with  the  characteristic  of  the  given 
logarithm  by  pointing  off  decimals  or  annexing  ciphers.  Thus 
the  natural  number  belonging  to  the  log.  4.370143  is  23450 ; 
"  «  "  "  1.538574  is  34.56. 

If  the  decimal  part  of  the  logarithm  can  not  be  exactly  found 
in  the  table,  look  for  the  nearest  less  logarithm,  and  take  out 


14  Trigoinometr'! 

the  four  figures  of  the  corresponding  natural  number  as  be- 
fore ;  the  additional  figures  may  be  obtained  by  means  of  the 
Proportional  Parts  at  the  bottom  of  the  page. 

Required  the  number  belonging  to  the  logarithm  4.368399. 

On  page  6,  we  find  the  next  less  logarithm  .368287. 

The  four  corresponding  figures  of  the  natural  number  are 
2335.  Their  logarithm  is  less  than  the  one  proposed  by  112. 
The  tabular  difference  is  186  ;  and,  by  referring  to  the  bottom 
of  page  6,  we  find  that,  with  a  difference  of  186,  the  figure 
corresponding  to  the  proportional  part  112  is  6.  Hence  the 
five  figures  of  the  natural  number  are  23356 ;  and,  since  the 
characteristic  of  the  proposed  logarithm  is  4,  these  five  figures 
are  all  integral. 

Required  the  number  belonging  to  logarithm  5.345678. 

The  next  less  logarithm  in  the  table  is  345570, 

Their  difference  is  108. 

The  first  four  figures  of  the  natural  number  are     2216. 

With  the  tabular  difference  196,  the  fifth  figure,  correspond- 
ing to  108,  is  seen  to  be  5,  with  a  remainder  of  10.  To  find 
the  sixth  figure  corresponding  to  this  remainder  10,  we  may 
multiply  it  by  10,  making  100,  and  search  for  100  in  the  same 
line  of  proportional  parts.  We  see  that  a  difference  of  100 
would  give  us  5  in  the  fifth  place  of  the  natural  number. 
Therefore,  a  difference  of  10  must  give  us  5  in  the  sixth  place 
of  the  natural  number.    Hence  the  required  number  is  221655. 

In  the  same  manner  we  find 

the  number  corresponding  to  log.  3.538672  is  3456.78  ; 
"  "  "       1.994605  is      98.7654; 

"  "  "       1.647817  is  .444444 

Multiplication  by  Logarithms. 
(11.)  According  to  Art.  3,  the  logarithm  of  the  product  ol 
two  or  more  factors  is  equal  to  the  sum  of  the  logarithms  of 
those  factors.     Hence,  for  multiplication  by  logarithms,  we 
have  the  following 

Rule. 
Add  the  logarithms  of  the  factors  ;  the  sum  will  be  the  log 
arithm  of  their  product. 

Ex.  1.  Req/iired  the  product  of  57.98  by  18. 


Logarithms.  lo 

The  logarithm  of  57.98  is  1.763278 

"  "  18  is  1.255273 

The  bgarithm  of  the  product  1043.64  is  3.018551 
Ex.  2.  Required  the  product  of  397.65  by  43.78. 

Ans.,  17409.117. 
Ex.  3.  Required  the  continued  product  of  54.32,  6.543,  and 
12.345. 

The  word  sum,  in  the  preceding  rule,  is  to  he  understood  in 
its  algebraic  sense  ;  therefore,  if  any  of  the  characteristics  of 
the  logarithms  are  negative,  we  must  take  the  difference  be- 
tween their  sum  and  that  of  the  positive  characteristics,  and 
prefix  the  sign  of  the  greater.  It  should  be  remembered  that 
the  decimal  part  of  the  logarithm  is  invariably  positive ;  hence 
that  which  is  carried  from  the  decimal  part  to  the  character- 
istic must  be  considered  po^sitive. 
Ex.  4.  Multiply  0.00563  b/  17. 

The  logarithm  of  0.00563  is  3.750508 
"  "  17      is  1.230449 

Product,  0.09571,  whose  logarithm  is  2.980957. 

Ex.  5.  Multiply  0.3854  by  0.0576.  Ans.  0.022199. 

Ex.  6.  Multiply  0.007853  by  0.00476. 

Ans.,  0.0000373b. 
Ex.  7.  Find  the  continued  product  of  11.35, 0.072,  and  0.017. 
^  (12.)  Negative  quantities  may  be  multiplied  by  means  of 
logarithms  in  the  same  manner  as  positive,  the  proper  sign 
being  prefixed  to  the  result  according  to  the  rules  of  Algebra. 
To  distinguish  the  negative  sign  of  a  natural  number  from  the 
negative  characteristic  of  a  logarithm,  we  append  the  letter  n 
to  the  logarithm  of  a  negative  factor.     Thus 

the  logarithm  of  —5Q  we  write  1.748188  n. 
Ex.  8.  Multiply  53.46  by  -29.47. 

The  logarithm  of     53.46  is  1.728029 
"  "  -29.47  is  1.469380  n. 

Product,  -1575.47,  log.  3.197409  n. 
Ex.  9.  Find  the, continued  product  of  372.1,  —.0054,  and 
-175.6. 

Ex.  10.  Find  the  continued  product  of  -0.137,  -7.689,  and 
-  0376. 


tfi  Trigonometry. 


A 


Division  by  Logarithms 


"(13.)  According  to  Art.  3,  the  logarithm  of  the  qaotient  ol 
one  number  divided  by  another  is  equal  to  the  difference  of 
the  logarithms  of  those  numbers.  Hence,  for  division  by  log. 
arithms,  we  have  the  following 

Rule. 

From  the  logarithm  of  the  dividend,  subtract  the  logarithm 
of  the  divisor ;  the  difference  will  he  the  logarithm  of  the 
quotient. 

Ex.  1.  Required  the  quotient  of  888.7  divided  by  42.24. 
The  logarithm  of  888.7  is  2.948755 
<'  "  42.24  is  1.625724 

The  quotient  is  21.039,  whose  log.  is  1.323031. 
Ex.  2.  Required  the  quotient  of  3807.6  divided  by  13.7. 

Ans.,  277.927. 
J     The  word  difference,  in  the  preceding  rule,  is  to  be  under- 
stood in  its  algebraic  sense ;  therefore,  if  the  characteristic  of 
one  of  the  logarithms  is  negative,  or  the  lower  one  is  greatei 
than  the  upper,  we  must  change  the  sign  of  the  subtrahend, 
and  proceed  as  in  addition.     If  unity  is  carried  from  the  deci- 
mal part,  this  must  be  considered  as  positive,  and  must  hp 
united  with  the  characteristic  before  its  sign  is  changed. 
Ex.  3.  Required  the  quotient  of  56.4  divided  by  0.00015. 
The  logarithm  of        56.4  is  1.75i279 
"  "  0.00015  is  4.176091 

The  quotient  is  376000,  whose  log.  is  5.575188. 
This  result  may  be  verified  in  the  same  way  as  subtraction 
in  common  arithmetic.  The  remainder,  added  to  the  subtra- 
hend, should  be  equal  to  the  minuend.  This  precaution  should 
always  be  observed  when  there  is  any  doubt  with  regard  to 
the  sign  of  the  result. 

Ex.  4.  Required  the  quotient  of  .8692  divided  by  42.258. 

Ans. 
Ex.  5.  Required  the  quotient  of  .74274  divided  by  .00928. 
Ex.  6.  Required  the  quotient  of  24.934  divided  by  .078541. 
Negative  quantities  may  be  divided  by  means  of  logarithms 


Logarithms.  17 

in  the  same  manner  as  positive,  the  pr  )per  sign  being  prefixed 
to  the  result  according  to  the  rules  of  Algebra. 
r     Ex.  7.  Required  the  quotient  of  -79.54  divided  by  0.08321 
C/  Ex.  8.  Required  the  quotient  of  -0.4753  divided  by  -36.74. 

y  Involution  by  Logarithms. 

(14.)  It  is  proved  in  Algebra,  Art.  340,  that  the  logarithm 
of  any  power  of  a  number  is  equal  to  the  logarithm  of  that 
number  multiplied  by  the  exponent  of  the  power.  Hence,  to 
involve  a  number  by  logarithms,  we  have  the  following 

Rule. 
Multiply  the  logarithm  of  the  number  by  the  exponent  of 
the  power  required. 

Ex.  1.  Required  the  square  of  428. 

The  logarithm  of  428  is  2.631444 

2 

Square,  183184,  log.  5.262888. 
Ex.  2.  Required  the  20th  power  of  1.06. 

The  logarithm  of  1.06  is  0.025306 

20 


20th  power,  3.2071,  log.  0.506120. 

Ex..  3.  Required  the  5th  power  of  2.846. 

It  should  be  remembered,  that  w^hat  is  carried  from  the  dec- 
imal part  of  the  logarithm  is  positive,  whether  the  characteris- 
tic is  positive  or  negative. 

Ex.  4.  Required  the  cube  of  .07654. 

The  logarithm  of  .07654  is  2.883888 

3 

Cube,  .0004484,  log.  4.651664. 
Ex.  5.  Required  the  fourth  power  of  0.09874. 
Ex.  6.  Required  the  seventh  power  of  0.8952. 

Evolution  by  Logarithms. 

(15.)  It  is  proved  in  Algebra,  Art.  341,  that  the  logarithm 
jf  any  root  of  a  number  is  equal  to  the  logarithm  of  that  num- 
ber divided  by  the  index  of  the  root.     Hence,  to  extract  the 
oot  of  a  number  by  logarithms,  we  have  the  following 

B 


18  Trigonometry. 

E/ULE. 

Divide  the  logarithm  of  the  number  by  the  index  of  th% 
root  required. 

Ex.  1.  Required  the  cu"be  root  of  482.38. 

The  logarithm  of  482.38  is  2.683389. 

Dividing  by  3,  we  have  0.894463,  which  corresponds  to 
7.842,  which  is  therefore  the  root  required. 
^  Ex.  2.  Required  the  100th  root  of  365. 

Ans.,  1.0608. 

When  the  characteristic  of  the  logarithm  is  negative,  and  is 
not  divisible  by  the  given  divisor,  we  may  increase  the  char- 
acteristic by  any  number  which  will  make  it  exactly  divisible, 
provided  we  prefix  an  equal  positive  number  to  the  decimal 
part  of  the  logarithm. 

Ex.  3.  Required  the  seventh  root  of  0.005846. 

The  logarithm  of  0.005846  is  3.766859,  which  may  be  writ- 
ten 7+4.766859. 

Dividing  by  7,  we  have  1.680980,  which  is  the  logarithm  of 
.4797,  which  is,  therefore,  the  root  required. 
•  .   This  result  may  be  verified  by  multiplying  1.680980  by  7 , 
the  result  will  be  found  to  be  3.766860. 
-    Ex.  4.  Required  the  fifth  root  of  0.08452. 
■i    Ex.  5.  Required  the  tenth  root  of  0.007815. 

Proportion  by  Logarithms. 
(16.)  The  fourth  term  of  a  proportion  is  found  by  multiply- 
ing together  the  second  and»  third  terms,  and  dividing  by  the 
first.     Hence,  to  find  the  fourth  term  of  a  proportion  by  loga 
rithms, 
.^;^  Add  the  logarithms  of  the  second  and  third  terms,  and  from 
their  sum  subtract  the  logarithm  of  the  first  term. 

Ex.  1.  Find  a  fourth  proportional  to  72.34,  2.519,  and  357.48 

Ans.,  12.448. 
,^  (17.)  When  one  logarithm  is  to  be  subtracted  from  anothej, 
it  may  be  more  convenient  to  convert  the  subtraction  into  an 
addition,  which  may  be  done  by  first  subtracting  the  given  log* 
arithm  from  10,  adding  the  difference  to  the  other  iogaiithni 
and  afterward  rejecting  the  10. 


Logarithms.  19 

"MPhe  difference  between  a  given  logarithm  and  10  is  called 
its  complement ;  and  this  is  easily  taken  from  the  table  by  be- 
ginning at  the  left  hand,  subtracting  each  figure  from  9,  ex- 
cept the  last  significant  figure  on  the  right,  which  must  be 
subtracted  from  10. 

i  To  subtract  one  logarithm  from  another  is  the  same  as  to 
add  its  complement,  and  then  reject  10  from  the  result.  For 
a—h\s,  equivalent  to  10— Z>+a— 10. 

To  work  a  proportion,  then,  by  logarithms,  we  must 
^    Add  the  complement  of  the  logarithm  of  the  first  term  to 
the  logarithms  of  the  second  and  third  terms. 

The  characteristic  must  afterward  be  diminished  by  10. 
Ex.  2.  Find  a  fourth  proportional  to  6853,  489,  and  38750. 
The  complement  of  the  logarithm  of  6853  is  6.164119 
The  logarithm  of  489  is  2.689309 

"  "  38750  is  4.588272 

The  fourth  term  is  2765,  whose  logarithm  is  3.441700. 
One  advantage  of  using  the  complement  of  the  first  term  in 
working  a  proportion  by  logarithrns  is,  that  it  enables  us  to 
exhibit  the  operation  in  a  more  compact  form. 
^  Ex.  3.  Find  a  fourth  proportional  to  73.84,  658.3,  and  4872. 

Ans. 
^x.  4.  Find  a  fcurth  proportional  to  5.745,  781.2,  and  54.27 


BOOK  IL 


PLANE  TRIGOI^iOMETRY. 

/][18.)  Trigonometry  is  the  iscience  ixrhuK  tntcchni,  hhw  to  dts 
termine  the  several  parts  of  a  triangle  from  having  certain 
parts  given. 

Plane  Trigonometry  treats  of  plane   triangles ;   Spherical 
y      Trigonometry  treats  of  spherical  triangles. 

J (19.)  The  circumference  of  every  circle  is  supposed  to  be 

/divided  into  360  equal  parts,  called  degrees  ;  each  degree  into 
60  minutes,  and  each  minute  into  60  seconds.  Degrees,  min- 
utes, and  seconds  are  designated  by  the  characters  °,  ',  ". 
Thus  23"  14'  35"  is  read  23  degrees,  14  minutes,  and  35  sec- 
onds. 

Since  an  angle  at  the  center  of  a  circle  is  measured  by  the 
^      arc  intercepted  by  its  sides,  a  right  angle  is  measured  by  90°, 
two  right  angles  by  180°,  and  four  right  angles  are  measured 
by  360°. 

(20.)  The  complement  of  an  arc  is  what  remains  after  sub- 
J    tracting  the  arc  from  90°-     Thus  the 
«c  DF  is  the  complement  of  AF. 
The  complement  of  25°  15'  is  64°  45'. 

In  general,  if  we  represent  any  arc 
by  A,  its  complement  is  90°— A. 
Hence,  if  an  arc  is  greater  than  90°, 
its  complement  must  be  negative. 
Thus,  the  complement  of  100°  15'  is 
—10°  15'.  Since  the  two  acute  an- 
gles of  a  right-angled  triangle  are  to- 
gether equal  to  a  right  angle,  each  of  them  must  be  the  com- 
plement of  the  other. 

(21.)  The  supplement  of  an  arc  is  what  remains  after  sub- 
tracting the  arc  from  180°.  Thus  the  arc  BDF  is  the  supple- 
ment of  the  arc  AF.  The  supplement  of  25°  15'  is  154°  45'. 
X  Tn  general,  if  we  represent  any  arc  by  A,  its  supplement  is 


P;,ANE    Trigonometry.  21 

180°— A.     Hence,  if  an  arc  is  greater  than  180°,  its  supple- 
ment must  be  negative.    Thus  the  supplement  of  200°  is  —20'. 
Since  in  every  triangle  the  sum  of  the  three  angles  is  180°, 
either  angle  is  the  supplement  of  the  sum  of  the  other  two. 
^    (22.)  The  sine  of  an  arc  is  the  perpendicular  let  fall  from 
one  extremity  of  the  arc  on  the  radius  passing'  through  the 
other  extremity.     Thus  FGr  is  the  sine  of  the  arc  AF,  or  of  the 
angle  ACF. 
>       Every  sine  is  half  the  chord  of  double  the  arc.     Thus  the 
sine  FG-  is  the  half  of  FH,  which  is  the  chord  of  the  arc  FAH, 
double  of  FA.     The  chord  which  subtends  the  sixth  part  of 
the  circumference,  or  the  chord  of  60°,  is  equal  to  the  radius 
(Geom.y  Prop.  lY.,  B.  VI.) ;  hence  the  sine  of  30°  is  equal  to 
half  of  the  radius. 
V  (23.)  The  versed  sine  of  an  arc  is  that  part  of  the  diameter 

^     intercepted  between  the  sine  and  the  arc.     Thus  GrA  is  the 

versed  sine  of  the  arc  AF. 
\        (24).  The  tangent  of  an  arc  is  the  line  which  touches  it  at 
one  extremity^  and  is  terminated  by  a  line  drawn  from  the 
center  through  the  other  extremity.     Thus  AI  is  the  tangent 
of  the  arc  AF,  or  of  the  angle  ACF. 
"^      (25.)  The  secant  of  an  arc  is  the  line  drawn  from  the  cen- 
ter of  the  circle  through  one  extremity  of  the  arc^  and  is  lim 
ited  by  the  tangent  drawn  through  the  other  extremity. 
Thus  CI  is  the  secant  of  the  arc  AF,  or  of  the  atigle  ACF. 
^^      (26.)  The  cosine  of  an  arc  is  the  sine  of  the  complement  of 
that  arc.     Thus  the  arc  DF,  being  the  complement  of  AF,  FK 
is  the  sine  of  the  arc  DF,  or  the  cosine  of  the  arc  AF. 
^^      The  cotangent  of  an  arc  is  the  tangent  of  the  complement 
of  that  arc.     Thus  DL  is  the  tangent  of  the  arc  DF,  or  the  co- 
tangent of  the  arc  AF. 

The  cosecant  of  an  arc  is  the  secant  of  the  complement  ol 
that  arc.  Thus  CL  is  the  secant  of  the  arc  DF,  or  the  cose- 
cant of  the  arc  AF. 

In  general,  if  we  represent  any  angle  by  A, 
cos.      A=sine    (90°-A). 
cot.      A=tang.  (90° -A), 
cosec.  A=sec.     (90°— A). 
Since,  in  a  right-angled  triangle,  either  of  the  acute  angle* 


^2 


Trigonometry., 


is  the  complement  of  the  other,  the  sine,  tangent,  and  secant 
i)f  one  of  these  angles  is  the  cosine,  cotangent,  and  cosecan 
of  the  other. 

(27.)  The  sine,  tangent,  and  secant  of  an  arc  are  equal  tc 
the  sine,  tangent,  and  secant  of  its  supplement.  Thus  FGr  is 
the  sine  of  the  arc  AF,  or  of  its  sup- 
plement, BDF.  Also,  AI,  the  tan- 
gent of  the  arc  AF,  is  equal  to  BM, 
the  tangent  of  the  arc  BDF.  And 
CI,  the  secant  of  the  arc  AF,  is  equal 
to  CM,  the  secant  of  the  arc  BDF. 

The  versed  sine  of  an  acute  angle, 
ACF,  is  equal  to  the  radius  minus 
the  cosine  CGr.  The  versed  sine  of 
an  obtuse  angle,  BCF,  is  equal  to  ra- 
dius plus  the  cosine  CGr ;  that  is,  to  BGr. 

(28.)  The  relations  of  the  sine,  cosine,  &c.,  to  each  other, 
may  he  derived  from  the  proportions  of  the  sides  of  similar 
triangles.  Thus  the  triangles  CG-F,  CAI,  CDL,  being  similar, 
we  have, 

1.  CGr  :  GrF  :  :  CA  :  AI ;  that  is,  representing  the  arc  by  A, 

and  the  radius  of  the  circle  by  R,  cos.  A  :  sin.  A  : :  R  :  tang.  A. 

A     ^  sin.  A 
Whence  tang.  A=- 


2.  Ca  :  :  CF  :  CA 


CI;  that  is,  cos.  A  :  R  : 

Whence  sec.  A 


cos.  A 
R  :  sec.  A. 
R'' 


cos.  A' 

aF  :  ca  : :  CD  :  DL ;  that  is,  sin.  A  :  cos.  A  :  :  R  :  cot.  A 

R  COS.  A 
sin.  A 


"Whence  cot.  A=- 


4    aF  :  CF  : :  CD  :  CL  ;  that  is,  sin.  A  :  R  : :  R  :  cosec.  A. 

Whence  cosec.  A=- r- 

sm.  A 

5.  AI ;  AC  : :  CD  :  DL ;  that  is,  tang.  A  :  R  : :  R  :  cot.  A. 

Whence  tanff.  A.= — ■ — r- 
°  cot.  A 

The  preceding  values  of  tangent  and  cotangent,  secant  and 

cosecant  will  be  frequently  referred  to  hereafter,  and  should 

be  catefuUy  committed  to  memory. 


Plane    Trigonometrv. 


23 


Also,  in  the  right-angled  triangle  CaF,  we  find  CGr'+aF'^ 
CF' ;  that  is,  sin.  'A+cos.  ''A=R' ;  or, 

The  square  of  the  sine  of  an  arc^  together  with  the  square 
if  its  cosine^  is  equal  to  the  square  of  the  radius. 

Hence  sin.  A=±  VR''— cos.  '^A. 
And  COS.     A=  ±  VR'*— sin.  'A. 
(29.)  A  table  of  natural  sines,  tangents,  &o.,  is  a  table  giv- 
ing the  lengths  of  those  lines  for  different  angles  in  a  circle 
whose  radius  is  unity. 

Thus,  if  we  describe  a  circle  with  a  radius  of  one  inch,  and 
divide  the  circumference  into  equal  parts  of  ten  degrees,  we 
fhall  find 

the  sine  of  10°  equals  0.174  inch ; 


20° 
30° 
40° 
50° 
60° 
70° 
80° 
90° 


0.342 
0.500 
0.643 
0.766 
0.866 
0.940 
0.985 
1.000 


If  we  draw  the  tangents  of  the  same  arcs,  we  shall  find 

the  tangent  of  10°  equals  0.176  inch ; 
a 

a 

a 

II 
li 
u 
II 
ii 
a 


20° 
30° 
40° 
45° 
50° 
60° 
70° 
80° 
90° 


0.364 
0.577 
0.839 
1.000 
1.192 
1.732 
2.747 
5.671 
infinite. 


Also,  if  we  draw  the  secants  of  the  same 
arcs,  we  shall  find  that 

the  secant  of  10°  equals  1.015  inch ; 
"  u         20°      "      1.064     " 

"  "         30°      "      1.155     " 

it  «         40°      a      1.305     " 


24  Trigonometry. 

the  secant  of  50°  equals  1.556  inch ; 
u   ^     u         60°      "      2.000     '' 
u   "     u         70°      u      2.924     " 
«         "         80°      "      5.759     " 
"  "         90°      "      infinite. 

In  tlie  accompanying  table,  pages  116-133,  the  sines,  co- 
sines,  tangents,  and  cotangents  are  given  for  every  minute  of 
the  quadrant  to  six  places  of  figures. 

(30.)  To  find  from  the  table  the  natural  sine,  cosine,  Sfc., 
of  an  arc  or  angle. 

If  a  sine  is  required,  look  for  the  degrees  at  the  top  of  the 
page,^and  for  the  minutes  on  the  left ;  then,  directly  under  the 
given  number  of  degrees  at  the  top  of  the  page,  and  opposite 
to  the  minutes  on  the  left,  will  be  found  the  sine  required. 
Since  the  radius  of  the  circle  is, supposed  to  be  unity,  the  sine 
ef  every  arc  below  90°  is  less  than  unity.  The  sines  are  ex- 
pressed in  decimal  parts  of  radius  ;  and,  although  the  decimal 
point  is  not  written  in  the  fable,  it  must  always  be  prefixed. 
As  the  first  two  figures  remain  the  same  for  a  great  many 
numbers  in  the  table,  they  are  only  inserted  for  every  ten  min- 
utes, and  the  vacant  places  must  be  supplied  from  the  two 
leading  figures  next  preceding      Thus,  on 

page  120,  the  sine  of  25°  11'  is  0.425516 ; 
page  126,    "        "       51°  34'  is  0.783332,  &o. 
The  tangents  are  found  in  a  similar  manner.     Thus 
the  tangent  of  31°  44'  is  0.618417  ; 
"  "         Q>6°  27'  is  2.18923. 

The  same  number  in  the  table  is  both  the  sine  of  an  arc  and 
the  cosine  of  its  complement.  The  degrees  for  the  cosines 
must  be  sought  at  the  bottom  of  the  page,  and  the  minutes  on 
the  right.     Thus, 

on  page  130,  the  cosine  of  16°  42'  is  0.957822  ; 
on  page  118,   "        >'         73°  17'  is  0.287639,  &c. 
The  cotangents  are  found  in  the  same  manner.     Thus 
the  cotangent  of  19°  16'  is  2.86089 ; 
"  ''  54°  53'  is  0.703246. 

It  is  not  necessary  to  extend  the  tables  beyond  a  quadrant, 
because  the  sine  of  an  angle  is  equal  to  that  of  its  supplemeni 
(Art.  27).     Thus 


Plane    Trigono  etry.  "23 

the  sine  of  143°  24'  is  0.596225  ; 

"  cosine        of  151°  23' is  0.877844; 
"  tangent      of  132°  36'  is  1.08749  ; 
"    cotangent  of  116°    7'  is  0.490256,  &o. 
(31.)  If  a  sine  is  required  for  an  arc  consisting  of  degrees, 
minutes,  and  seconds^  we  must  make  an  allowance  for  the  sec- 
onds in  the  same  manner  as  was  directed  in  the  case  of  loga- 
rithms, Art.  7  ;  for,  within  certain  limits,  the  differences  of  the 
sines  are  proportional  to  the  differences  of  the  corresponding 
arcs.     Thus 

the  sine  of  34°  25'  is  .565207 ; 
"  "  34°  26'  is  .565447. 
The  difference  of  the  sines  corresponding  to  one  minute  of 
arc,  or  60  seconds,  is  .000240.  The  proportional  part  for  1 '  is 
found  by  dividing  the  tabular  difference  by  60,  and  the  quo- 
tient, .000004,  is  placed  at  the  bottom  of  page  122,  in  the  col- 
umn headed  34°.  The  correction  for  any  number  of  seconds 
will  be  found  by  multiplying  the  proportional  part  for  1"  by 
the  number  of  seconds. 

Required  the  natural  sine  of  34°  25'  37". 
The  proportional  part  for  1",  being  multiplied  by  37,  becomes 
148,  which  is  the  correction  for  37".     Adding  this  to  the  sine 
of  34°  25',  we  find 

the  sine  of  34°  25'  37"  is  .^5^66. 
Since  the  proportional  part  for  1"  is  given  to  hundredths  of  a 
unit  in  the  sixth  place  of  figures,  after  we  have  multiplied  by 
the  given  number  of  seconds,  we  must  reject  the  last  two  fig- 
ures of  the  product. 

In  the  same  manner  we  find 

the  cosine  of  5%''  34'  28"  is  .550853. 
it  will  be  observed,  that  since  the  cosines  decrease  while 
the  arcs  increase,  the  correction  for  the  28"  is  to  be  subtracted 
from  ^he  cosine  of  56°  34'. 
In  the  same  manner  we  find 

the  natural  sine  of  27°  17'  12"  is  0.458443  ; 

"         "      cosine        of  45°  23'  23"  is  0.702281 ; 
"         "      tangent     of  63°  32'  34"  is  2.00945 : 
**         "      cotangent  of  81°  48'  bQ"  is  0.143825 


26  .       Trigonometry. 

(32.)  To  find  the  number  of  decrees,  minutes^  and  seconds 
lelonging'  to  a  given  sine  or  tangent. 

If  the  given  sine  or  tangent  is  found  exactly  in  the  table, 
the  corresponding  degrees  will  be  found  at  the  top  of  the  page, 
and  the  minutes  on  the  left  hand.  But  when  the  given  num- 
ber is  not  found  exactly  in  the  table,  look  for  the  sine  or  tan- 
gent which  is  next  less  than  the  proposed  one,  and  take  out 
the  corresponding  degrees  and  minutes.  Find,  also,  the  dif- 
ference between  this  tabular  number  and  the  number  proposed, 
and  divide  it  by  the  proportional  part  for  1"  found  at  the  bot 
tom  of  the  page  ;  the  quotient  will  be  the  required  number  of 
seconds. 

Required  the  arc  whose  sine  is  .750000. 

The  next  less  sine  in  the  table  is  .749919,  the  arc  correspond- 
ing to  which  is  48°  35'.  The  difference  between  this  sine  and 
that  proposed  is  81,  which,  divided  by  3.21,  gives  25.  Hence 
the  required  arc  is  48°  35'  25". 

In  the  same  manner  we  find 

the  arc  whose  tangent  is  2.00000  is  63°  26'  6". 

If  a  cosine  or  cotangent  is  required,  we  must  look  for  the 
number,  in  the  table  which  is  next  greater  than  the  one  pro- 
posed, and  then  proceed  as  for  a  sine  or  tangent.     Thus 
the  arc  whose  cosine        is    .40000  is  m^"  25'  18"  ; 
''<'     "        «'      cotangent  is  1.99468  is  26°  37'  34". 

(33.)  On  pages  134-5  will  be  found  a  table  of  natural  se- 
cants for  every  ten  minutes  of  the  quadrant,  carried  to  seven 
places  of  figures.  The  degrees  are  arranged  in  .order  in  the 
first  vertical  column  on  the.  left,  and  the  minutes  at  the  top 
of  the  page.     Thus 

the  secant  of  21°  20'  is  1.073561 ; 
''         "         81°  §0'  is  7.039622. 

If  a  secant  is  required  for  a  number  of  minutes  not  given  in 
the  table,  the  correction  for  the  odd  minutes  may  be  foiftid  by 
means  of  the  last  vertical  column  on  the  right,  which  shows 
the  proportional  part  for  one  minute. 

Let  it  be  required  to  find  the  secant  of  30°  33' 
The  secant  of  30°  30'  is  1.160592. 

The  correction  for  1'  is  198.9,  which,  multiplied  by  3,  be- 


,  Plane   Trigonometry.  27 

3omes  597.     Adding  this  to  the  number  before  found,  we  ob- 
tain 1.161189. 

For  a  cosecant,  the  degrees  must  be  sought  in  the  right- 
hand  vertical  column,  and  the  minutes  at  the  bottom  of  th*» 
page.     Thus 

the  cosecant  of  47°  40'  is  1.352742; 
"  "         38°  33'  is  1.604626. 

(34.)  When  the  natural  sines,  tangents,  &c.,  are  used  in  pro- 
portions, it  is  necessary  to  perform  the  tedious  operations  of 
multiplication  and  division.  It  is,  therefore,  generally  prefer- 
able to  employ  the  logarithms  of  the  sines ;  and,  for  conven- 
ience, these  numbers  are  arranged  in  a  separate  table,  called 
Jogarithmic  sinesy  &c.     Thus 

the  natural  sine  of  14°  30'  is  0.250380. 

Its  logarithm,  found  from  page  6,  is  1.398600. 

The  characteristic  of  the  logarithm  is  negative,  as  must  be 
the  case  with  all  the  sines,  since  they  are  less  than  unity.  To 
avoid  the  introduction  of  negative  numbers  in  the  table,  we  in- 
crease the  characteristic  by  10,  making  9.398600,  and  this  is 
the  number  found  on  page  38  for  the  logarithmic  sine  of  14° 
30'.  The  radius  of  the  table  of  logarithmic  sines  is  therefore, 
properly,  10,000,000,000,  whose  logarithm  is  10. 

(35.)  The  accompanying  table  contains  the  logarithmic  sines 
and  tangents  for  every  ten  seconds  of  the  quadrant.  The  de- 
grees and  seconds  are  placed  at  the  top  of  the  page,  and  the 
minutes  in  the  left  vertical  column.  After  the  first  two  de- 
grees, the  three  leading  figures  in  the  table  of  sines  are  only 
given  in  the  column  headed  0",  and  are  to  be  prefixed  to  the 
numbers  in  the  other  columns,  as  in  the  table  of  logarithms  of 
numbers.  Also,  where  the  leading  figures  change,  this  change 
is  indicated  by  dots,  as  in  the  former  table.  The  correction 
for  any  number  of  seconds  less  than  10  is  given  at  the  bottom 
of  the  page. 

(36.)  To  find  the  logarithmic  sine  or  tangent  of  a  given 
arc. 

Look  for  the  degrees  at  the  top  of  the  page,  the  minutes  on 
the  left  hand,  and  the  next  less  number  of  seconds  at  the  top ; 
then,  under  the  seconds,  and  opposite  to  the  minutes,  will  be 
tound  four  figures,  to  which  the  three  leading  figures  are  to  be 


28  Trigonometry. 

prefixed  from  the  column  headed  0^^ ;  to  this  add  the  propor. 
tional  part  for  the  odd  seconds  at  the  bottom  of  the  page. 
Required  the  logarithmic  sine  of  24"  27'  34". 

The  logarithmic  sine  of  24°  27'  30"  is  9.617033 
Proportional  part  for  4"  is  18 

Logarithmic  sine  of  24°  27'  34"  is       9.617051. 
Required  the  logarithmic  tangent  of  73°  35'  43". 
The  logarithmic  tangent  73°  35'  40"  is  10.531031 
Proportional  part  for  3"  is  23 

Logarithmic  tangent  of  73°  35'  43"  is    10.531054. 
When  a  cosine  is  required,  the  degrees  and  seconds  must  bo 
sought  at  the  bottom  of  the  page,  and  the  minutes  on  the  right, 
and  the  correction  for  the  odd  seconds  must  be  subtracted  from 
the  number  in  the  table. 

Required  the  logarithmic  cosine  of  59°  33'  47". 

The  logarithmic  cosine  of  59°  33'  40"  is  9.704682 
Proportional  part  for  7"  is  25 

Logarithmic  cosine  of  59°  33'  47"  is        9.704657. 

So,  also,  the  logarithmic  cotangent  of  37°  27'  14"  is  found 
to  be  10.115744. 

It  will  be  observed  that  for  the  cosines  and  cotangents,  the 
seconds  are  numbered  from  10"  to  60",  so  that  if  it  is  re- 
quired to  find  the  cosine  of  25°  25'  0"  we  must  look  for  25° 
24'  60" ;  and  so,  also,  for  the  cotangents. 

(37.)  The  proportional  parts  given  at  the  bottom  of  each 
page  correspond  to  the  degrees  at  the  top  of  the  page,  in- 
creased by  30',  and  are  not  strictly  applicable  to  any  other 
number  of  minutes ;  nevertheless,  the  differences  of  the  sines 
change  so  slowly,  except  near  the  commencement  of  the  quad- 
rant, that  the  error  resulting  from  using  these  numbers  for 
every  part  of  the  page  will  seldom  exceed  a  unit  in  the  sixth 
decimal  place.  For  the  first  two  degrees,  the  differences 
change  so  rapidly  that  the  proportional  part  for  1"  is  given  for 
each  minute  in  the  right-hand  column  of  the  page.  The  cor- 
rection for  any  number  of  seconds  less  than  ten  w  ill  be  found 
by  multiplying  the  proportional  part  for  1"  by  the  given  num- 
ber  of  seconds. 

Required  the  logarithmic  sine  of  1°  17'  33". 


Plane    Trigonometry.  29 

The  logarithmic  sine  of  1°  17'  30"  is  8.352991. 

The  correction  for  3"  is  found  by  multiplying  93.4  by  3, 
which  gives  280.  Adding  this  to  the  above  tabular  number, 
we  obtain  for 

the  sine  of  1°  17'  33",  8.353271. 

A  similar  method  may  be  employed  for  several  of  the  first 
degrees  of  the  quadrant,  if  the  proportional  parts  at  the  bottom 
of  the  page  are  not  thought  sufficiently  precise.  This  correc- 
tion may,  however,  be  obtained  pretty  nearly  by  inspection, 
from  comparing  the  proportional  parts  for  two  successive  de- 
grees. Thus,  on  page  26,  the  correction  for  1",  corresponding 
to  the  sine  of  2°  30',  is  48 ;  the  correction  for  1",  correspond- 
ing to  the  sine  of  3°  30',  is  34.  Hence  the  correction  for  1", 
corresponding  to  the  sine  of  3°  0',  must  be  about  41 ;  and,  in 
the  same  manner,  we  may  proceed  for  any  other  part  of  the 
table. 

(38.)  Near  the  close  of  the  quadrant,  the  tangents  vary  so 
rapidly  that  the  same  arrangement  of  the  table  is  adopted  as 
for  the  commencement  of  the  quadrant.  For  the  last,  as  well 
as  the  first  two  degrees  of  the  quadrant,  the  proportional  part 
to  1"  is  given  for  each  minute  separately.  These  proportional 
parts  are  computed  for  the  minutes  placed  opposite  to  them, 
increased  by  30",  and  are  not  strictly  applicable  to  any  other 
number  of  seconds ;  nevertheless,  the  differences  for  the  most 
part  change  so  slowly,  that  the  error  resulting  from  using  these 
numbers  for  every  part*  of  the  same  horizontal  line  is  quite 
small.  Whc  ii  great  accuracy  is  required,  the  table  on  page  114 
may  be  employed  for  arcs  near  the  limits  of  the  quadrant.  This 
table  furnishes  the  differences  between  the  logarithmic  sines 
and  the  logarithms  of  the  arcs  expressed  in  seconds.     Thus 

the  logarithmic  sine  of  0°  5',  from  page  22,  is  7.162696 

the  logarithm  of  300"  (=5')  is  2.477121 

the  difference  is  4.685575. 

This  is  the  number  found  on  page  114,  under  the  heading 
log',  sine  A— log:  A",  opposite  to  5  min. ;  and,  in  a  similar  man- 
ner, the  other  numbers  in  the  same  column  are  obtained.  These 
Qumbers  vary  quite  slowly  for  two  degrees  ;  and  hence,  to  find 
the  logarithmic  sine  of  an  arc  less  than  two  degrees,  we  havn 


30  Trigonometry. 

but  to  add  the  logarithm  of  the  arc  expressed  in  seconds  to  thft 
appropriate  number  found  in  this  table. 
Required  the  logarithmic,  sine  of  0°  7'  22". 

Tabular  number  from  page  114,  4.685575 
The  logarithm  of  442"  is  2.645422 

Logarithmic  sine  of  0°  7'  22"  is  7.330997.  • 
The  logarithmic  tangent  of  an  arc  less  than  two  degrees  is 
found  in  a  similar  manner. 

Required  the  logarithmic  tangent  of  0°  27'  36". 

Tabular  number  from  page  114,  4.685584 

The  logarithm  of  1656"  is  3.219060 

Logarithmic  tangent  of  0°  27'  36"  is  7.904644. 
The  column  headed  log.  cot.  A+log.  A",  is  found  by  adding 
the  logarithmic  cotangent  to  the  logarithm  of  the  arc  expressed 
in  seconds.  Hence,  to  find  the  logarithmic  cotangent  of  an  aro 
less  than  two  degrees,  we  must  subtract  from  the  tabular  num- 
ber the  logarithm  of  the  arc  in  seconds. 

Required  the  logarithmic  cotangent  of  0°  27'  36". 
Tabular  number  from  page  114,  15.314416 

The  logarithm  of  1656"  is  3.219060 

Logarithmic  cotangent  of  0°  27'  36"  is  12.095356. 
The  same  method  will,  of  course  <  furnish  cosines  and  cotan 
gents  of  arcs  near  90°. 

(39.)  The  secants  and  cosecants  are  omitted  in  this  table, 
since  they  are  easily  derived  from  the  cosines  and  sines.     We 

R=^ 
have  found,  Art.  28,  secant  = — -. — ;  or,  taking  the  logarithms, 

log.  secant  =2.  log.  R— log.  cosine 

=20— log.  cosine. 

Also,  cosecant  =  -: — , 

sme 

or  log.  cosecant       =20— log.  sine.     That  is. 

The  logarithmic  secant  is  found  by  subtracting  the  loga- 
rithmic  cosine  from  20 ;  and  the  logarithmic  cosecant  is  found 
by  subtracting  the  logarithmic  sine  from  20. 

Thus  we  have  found  the  logarithmic  sine  of  24°  27'  34'  to 
be  9.617051. 

Hence  the  logarithmic  cosecant  of  24°  27'  34"  is  10.382949 


Plane    Trigonometry.  31 

The  logarithmic  cosine  of  54°  12'  40"  is    9.767008. 

Hence  the  logarithmic  secant  of  54°  12'  40"  is  10.232992. 

(40.)  To  find  the  arc  corresponding  to  a  given  logarithmic 
sine  or  tangent. 

If  the  given  number  is  found  exactly  in  the  table,  the  cor- 
responding degrees  and  seconds  will  be  found  at  the  top  of  the 
page,  and  the  minutes  on  the  left.  But  when  the  given  num- 
ber is  not  found  exactly  in  the  table,  look  for  the  sine  or  tan- 
gent which  is  next  less  than  the  proposed  one,  and  take  out 
the  corresponding  degrees,  minutes,  and  seconds.  Find,  also, 
the  difference  between  this  tabular  number  and  the  number 
proposed,  and  corresponding  to  this  difference,  at  the  bottom 
of  the  page,  will  be  found  a  certain  number  of  seconds  which 
is  to  be  added  to  the  arc  before  found. 

Required  the  arc  corresponding  to  the  logarithmic  sine 
9.750000. 

The  next  less  sine  in  the  table  is  9.749987. 

The  arc  corresponding  to  which  is  34°  13'  0". 

The  difference  between  its  sine  and  the  one  proposed  is  13, 
corresponding  to  which,  at  the  bottom  of  the  page,  we  find  4'' 
nearly.     Hence  the  required  arc  is  34°  13'  4". 

In  the  same  manner,  we  find  the  arc  corresponding  to  loga- 
rithmic tangent  10.250000  to  be  60°  38'  57". 

When  the  arc  falls  within  the  first  two  degrees  of  the  quad- 
rant, the  odd  seconds  may  be  found  by  dividing  the  difference 
between  the  tabular  number  and  the  one  proposed,  by  the  pro- 
portional part  for  1".  We  thus  find  the  arc  corresponding  to 
logarithmic  sine  8.400000  to  be  1°  26'  22"  nearly. 

We  may  employ  the  same  method  for  the  last  two  degrees 
of  the  quadrant  when  a  tangent  is  given  ;  but  near  the  limits 
of  the  quadrant  it  is  better  to  employ  the  auxiliary  table  on 
page  114.  The  tabular  number  on  page  114  is  equal  to  log. 
sin.  A— log.  A".  Hence  log.  sin.  A—  tabular  number  =log. 
A"  ;  that  is,  if  we  subtract  the  corresponding  tabular  number 
on  page  114,  from  the  given  logarithmic  sine,  the  remainder 
will  be  the  logarithm  of  the  arc  expressed  in  seconds. 

Required  the  arc  corresponding  to  logarithmic  sine  7.000000. 

We  see,  from  page  22,  that  the  arc  must  be  nearly  3' ;  tho 
corresponding  tabular  number  on  page  114  is  4.685575 


32  Trigonometry. 

The  difference  is  2.314425, 
which  is  the  logarithm  of  206."265. 

Hence  the  required  arc  is  3'  26. "265. 

Required  the  arc  corresponding  to  log.  sine  8.000000. 

We  see  from  page  22,  that  the  arc  is  about  34'.  The  coi- 
responding  tabular  number  from  page  114  is  4.685568,  which, 
subtracted  from  8.000000,  leaves  3.314432,  which  is  the  log- 
arithm of  2062. "68.     Hence  the  required  arc  is 

34'  22."68. 

In  the  same  manner,  we  find  the  arc  corresponding  to  loga 
rithmic  tangent  8.184608  to  be  0°  52'  35". 

^  SOLUTIONS  OF  RIGHT-ANGLED  TRIANGLES. 

Theorem  I. 

(41.)  In  any  right-angled  triangle^  radius  is  to  the  hypoth- 
enuse  as  the  sine  of  either  acute  angle  is  to  the  opposite  side, 
or  the  cosine  of  either  acute  angle  to  the  adjacent  side. 

Let  the  triangle  CAB  be  right  angled 
at  A,  then  will 

R  :  CB  :  :  sin.  C  :  BA  : :  cos.  C  :  CA  y 

From  the  point  C  as  a  center,  with  a 
radius  equal  to  the  radius  of  the  tables,   C  F  D 

describe  the  arc  DE,  and  on  AC  let  fall  thjb  perpendicular  SF 
Then  EF  will  be  the  sine,  and  CF  the  cosine  of  the  angle  C. 
Because  the  triangles  CAB,  CFE  are  similar,  we  have 

CE  :  CB  :  :  EF  :  BA, 
or  R  :  CB  :  :  sin.  C  :  BA. 

Also,  CE  :  CB  :  :  CF  :  CA, 

or  R  :  CB  :  :  cos.  C  :  CA. 

Theorem  H. 

(42.)  In  any  right-angled  triangle,  radius  is  to  either  side 
as  the  tangent  of  the  adjacent  acute  angle  is  to  the  opposite 
side,  or  the  secant  of  the  same  angle  to  the  hypothenuse. 

Let  the  triangle  CAB  be  right  angled 
at  A,  then  will 

R  :  CA  : :  tang.  C  :  AB  : :  sec.  C  :  CB. 

From  the  point  C  as  a  center,  with  a 
^-adius  equal  to  the  radius  of  the  tables, 


cant  of  the 

angle 

5  C. 

Bec! 

similar,  we 

have 

CD: 

;CA 

or 

R  : 

:CA 

Also, 

CD: 

;  CA 

Plane    Trigonometry.  33 

descrihe  the -arc  DE,  and  from  the  point  D  draw  DF  perpen- 
dicular to  CA.     Then  DF  will  he  the  tangent,  and  CF  the  se- 
ise the  triangles  CAB,  CDF  are 
:  DF  :  AB, 
:  tang.  C  :  AB. 
:  CF  :  CB, 
or  R  :  CA  :  :  sec.  C  :  CB. 

(43.)  In  every  plane  triangle  there  are  six  parts  :  three  sides 
and  three  angles.  Of  these,  any  three  heing  given,  provided 
one  of  them  is  a  side,  the  others  may  be  determined.  In  a 
right-angled  triangle,  one  of  the  six  parts,  viz.,  the  right  angle, 
is  always  given ;  and  if  one  of  the  acute  angles  is  given,  the 
other  is,  of  course,  known.  Hence  the  number  of  parts  to  be 
considered  in  a  right-angled  triangle  is  reduced  to  four,  any 
two  of  which  being  given,  the  others  may  be  found. 

It  is  desirable  to  have  appropriate  names  by  which  to  des- 
ignate each  of  the  parts  of  a  triangle.  One  of  the  sides  ad- 
jacent to  the  right  angle  being  called  the  base,  the  other  side 
adjacent  to  the  right  angle  may  be  called  the  perpendicular. 
The  three  sides  will  then  be  called  the  hypothenUse,  base,  and 
perpendicular.  The  base  and,  perpendicular  are  sometimes 
called  the  legs  of  the  triangle.  Of  the  two  acute  angles,  that 
which  is  adjacent  to  the  base  may  be  called  the  angle  at  the 
base,  and  the  other  the  angle  at  the  perpendicular. 

We  may,  therefore,  have  four  cases,  according  as  there  are 
given, 

1.  The  hypothenuse  and  the  angles  ;  yi 

2.  The  hypothenuse  and  a  leg ; 

3.  One  leg  and  the  angles  ;  or, 

4.  The  two  legs. 

All  of  these  cases  may  be  solved  by  the  two  preceding  theo- 
rems. 

Case  I. 
(44.)  Given  the  hypothenuse  and  the  angles,  to  find  the  base 
and  perpendicular. 

This  case  is  solved  by  Theorem  I. 

Radius  :  hypothenuse  >i  sine  of  the  angle  at  the  base  :  per- 
pendicular ; 

.  : :  cosine  of  the  angle  at  the  base  :  basr  - 
C 


34  Trigonometry. 

Ex.  1.  G-iven  the  hypothenuse  275,  and  the  angle  at  the  haso 
57°  23',  to  find  the  base  and  perpendicular. 
The  natural  sine  of  57°  23'  is  .842296 ; 
"  •         cosine        ^'  .539016. 

Hence  1  :  275  :  :  .842296  :  231.631=AB. 
1  :  275  :  :  .539016  :  148.229=AC. 
The  computation  is  here  made  by  natural 
numbers.     If  we  work  the  proportion  by  loga- 
rithms, we  shall  have 

Radius,  10.000000 

Is  to  the  hypothenuse  275  2.439333 

As  the  sine  of  C  57°  23'  9.925465 

To  the  perpendicular  231.63  2.364798. 

Also,  Radius,  10.000000 

Is  to  the  hypothenuse  275  2.439333 

As  the  cosine  of  C  57°  23'  9.731602 

To  the  base  148.23  2.170935. 

Ex.  2.  Griven  the  hypothenuse  67.43,  and  the  angle  at  the 
perpendicular  38°  43',  to  find  the  base  and  perpendicular. 
Ans.  The  base  is  42.175,  and  perpendicular  52.612. 
The  student  should  work  this  and  the  following  examples 
both  by  natural  numbers  and  by  logarithms,  until  he  has  made 
himself  perfectly  familiar  with  both  methods.  He  may  then 
employ  either  method,  as  may  appear  to  him  most  expeditious 

Case  II. 

(45.)  Given  the  hypothenuse  and  one  leg.,  to  find  the  angles 
and  the  other  leg. 

This  case  is  solved  by  Theorem  I. 

Hypothenuse  :  radius  : ;  base  :  cosine  of  the  angle  at  the  base. 
Radius  :  hypothenuse  : :  sine  of  the  angle  at  the  base  : 
perpendicular. 

When  the  perpendicular  is  given,  perpendicular  must  be 
substituted  for  base  in  this  proportion. 

Ex.  1.  G-iven  the  hypothenuse  54.32,  and  the  base  32.11,  to 
find  the  angles  and  the  perpendicular. 

By  natural  numbers,  we  have 


Planf    Trigonometry.  35 

54.32  :  1  : :  32.11  :  .591127,  which  is  the  cosine  of  53°  45' 

47",  the  angl  *>  at  the  base. 

Also,  1  :  54.32  : :  .806580  :  43.813=the  perpendicular. 

The  computation  may  be  performed  more  expeditiously  by 
logarithms,  as  in  the  former  case. 

Ex.  2.  Griven  the  hypbthenuse  332.49,  and  the  perpendicu- 
lar 98.399,  to  find  the  angles  and  the  base. 

Ans.  The  angles  are  17°  12'  51"  and  72°  47'  9";  the  base. 
317.6. 

Case  III. 

(46.)  Given  one  leg'  and  the  angles,  to  find  the  other  leg 
and  hypothenuse. 

This  case  is  solved  by  Theorem  11. 
Radius  :  base  : :  tangent  of  the  angle  at  the  base  :  the  perpen 
dicular. 

: :  secant  of  the  angle  at  the  base  :  hypothenuse. 

When  the  perpendicplar  is  given,  perpendicular  must  be 
substituted  for  base  in  Ithis  proportion. 

Ex.  1.  Griven  the  base  222,  and  the  angle  at  th^base  25°  15', 
to  find  the  perpendicular  and  hypothenuse. 

By  natural  numbers,  we  have 

1  :  222  :  :    .471631  :  104.70,  perpendicular ; 
:  :  1.105638  :  245.45,  hypothenuse. 

The  computation  should  also  be  performed  by  logarithms,  as 
in  Case  I. 

Ex.  2.  Griven  the  perpendicular  125,  and  the  angle  at  the 
perpendicular  51°  19',  to  find  the  hypothenuse  and  base. 

Ans.  Hypothenuse,  199.99 ;  base,  156.12. 

Case  IY. 

(47.)  Given  the  two  legs,  to  find  the  angles  and  hypothenuse. 

This  case  is  solved  by  Theorem  II. 
Base  :  radius  :  -.perpendicular  :  tangent  of  the  angle  at  the  base. 
Radius :  base : :  secant  of  the  angle  at  the  base  :  hypothenuse. 

Ex.  1.  Griven  the  base  123,  and  perpendicular  765,  to  find 
the  angles  and  hypothenuse. 

By  natural  numbers,  we  have 

123  :  1  :  :  765  :  6.219512,  whicK  is  the  tangent  of  80°  51' 
57",  the  angle  at  the  base. 


36  Trigonometry. 

1  :  123  :  :  6.299338  :  774.82,  hypothenuse. 

The  computation  may  also  be  made  by  logarithms,  as  in 
Case  T. 

Ex.  2.  G-iven  the  base  53,  and  perpendicular  67,  to  find  the 
angles  and  hypothenuse. 

Ans.  The  angles  are  51°  39'  16"  and  38°  20'  44"  ;  hypothe- 
nuse,  85.428. 

Examples  for  Practice. 

1.  Griven  the  base  777,  and  perpendicular  345,  to  find  the 
hypothenuse  and  angles. 

This  example,  it  will  be  seen,  falls  under  Case  IV. 

2.  Griven  the  hypothenuse  324,  and  the  angle  at  the  base 
48°  17',  to  find  the  base  and  perpendicular. 

3.  Griven  the  perpendicular  543,  and  the  angle  at  the  base 
72°  45',  to  find  the  hypothenuse  and  base. 

4.  Griven  the  hypothenuse  666,  and  base  432,  to  find  the  an- 
gles and  perpendicular. 

5.  Griven  the  base  634,  and  the  angle  at  the  base  53°  27',  to 
find  the  hypothenuse  and  perpendicular. 

6.  Griven  the  hypothenuse  1234,  and  perpendicular  555,  to 
find  the  base  and  angles. 

(48.)  "When  two  sides  of  a  right-angled  triangle  are  given, 
the  third  may  be  found  by  means  of  the  property  that  the 
square  of  the  hypothenuse  is  equal  to  the  sum  of  the  squares 
of  the  other  two  sides. 

Hence,  representing  the  hypothenuse,  base,  and  perpendicu 
|p.r  by  the  initial  letters  of  these  words,  we  have 

h=V¥+f',  b=y/T^;  p=Vli^^\ 
Ex.  1.  If  the  base  is  2720,  and  the  perpendicular  3104,  what 

is  the  hypothenuse  ?  Ans.,  4127.1. 

Ex.  2.  If  the  hypothenuse  is  514,  and  the  perpendicular  432, 

what  is  the  base  ? 


V 


SOLUTIONS  OF  OBLIQUE-ANGLED  TRIANGLES. 
Theorem  I. 

(49.)  In  any  plane  triangle,  the  sines  of  the  angles  are 
fjrovortional  to  the  opposite  sides. 


nj 


Plane   Trigonometry. 


37 


Let  ABC  be  any  triangle,  and  from  one 
Df  its  angles,  as  C,  let  CD  be  drawn  per- 
pendicular to  AB.      Then,  because  the         /   -  i  .♦  \ 
triangle  ACD  is  right  angled  at  D,  we      ^        \\\ 
have  *  -Ai*  '   ^  \ 

R  :  sin.  A  : :  AC  :  CD ;  whence  RxCi)=^sin.  AxAC. 

For  the  same  reason, 
R  :  sin.  B  : :  BC  :  CD ;  whence  RxCD=sin.  BxBC. 

Therefore,  sin.  AxAC=sin.  BxBC, 

or  sin.  A  :  sin.  B  :  :  BC  :  AC. 


Theorem  II. 

(50.)  In  any  plane  triangle^  the  sum  of  any  tioo  sides  is  to 
their  difference,  as  the  tangent  of  half  the  sum  of  the  opposite 
angles  is  to  the  tangent  of  half  their  difference. 

Let  ABC  be  any  triangle  ;  then  will 

A+B  A-B 

tang. 


CB+CA :  CB~CA  : 


tang. 


2      o-     2 

Produce  AC  to  D,  making  CD  equal  to  CB,  and  join  DU. 
Take  CE  equal  to  CA,  draw  AE,  and  produce  it  to  F.  Then 
AD  is  the  sum  of  CB  and  CA,  and  BE  is  their  differ en^ce. 

The  sum  of  the  two  angles  CAE,  CEA,  is  equal  to  the  sum 
of  CAB,  CBA,  each 'being  the  supplement  of  ACB  {Geom., 
Prop.  27,  B.  I.).     But,  since  CA  is  equal  j^ 

to  CE,  the  angle  CAE  is  equal  to  the  an-  /"• 

gle  CEA;  therefore,  CAE  is  the  half 
sum  of  the  angles  CAB,  CBA.  Also,  if 
from  the  greater  of  the  two  angles  CAB, 
CBA,  there  be  taken  their  half  sum,' the 
remainder,  FAB,  will  be  their  half  differ- 
ence (Algebra,  p.  68). 

Since  CD  is  equal  to  CB,  the  angle  ADF 
is  equal  to  the  angle  EBF ;  also,  the  an- 
gle CAP  is  equal  to  AEC,  which  is  equal 
to  the  vertical  angle  BEF.  Therefore,  the  two  triangles  DAF, 
BEF,  are  mutually  equiangular ;  hence  the  two  angles  at  F 
are  equal,  and  AF  is  perpendicular  to  DB.  If,  then,  AF  be 
made  radius,  DF  will  be  the  tangent  of  DAF,  and  BF  will  be 
the  tangent  of  BAF.     But,  by  similar  triangles,  we  have 


38  Trkjonometry. 

AD  :  BE  :  :  DF  :  BF ;  that  is, 
OB-f  CA  :  CB-CA  : :  tang.  ^±?  :  tang.  ^^, 


Theorem  III. 

(51.)  If  from  any  angle  of  a  triangle  a  perpendicular  bt 
drawn  to  the  opposite  side  or  base,  the  whole  base  loill  be  to 
the  sum  of  the  other  two  sides,  as  the  difference  of  those  two 
sides  is  to  the  difference  of  the  segments  of  the  base. 

For  demonstration,  see  G-eometry,  Prop.  31,  Cor.,  B.  IV. 

(52).  In  every  plane  triangle,  three  parts  must  be  given  to 
enable  us  to  determine  the  others  ;  and  of  the  given  parts,  one, 
at  least,  must  be  a  side.  For  if  the  angles  only  are  given, 
these  might  belong  to  an  infinite  number  of  different  triangle? 
In  solving  oblique-angled  triangles,  four  different  ca?>«.s  may 
therefore  be  presented.     There  may  be  given, 

1.  Two  angles  and  a  side ; 

2.  Two  sides  and  an  angle  opposite  one  of  them  ; 

3.  Two  sides  and  the  included  angle ;  or, 

4.  The  three  sides. 

We  shall  represent  the  three  angles  of  the  proposed  triangle 
by  A,  B,  C,  and  the  sides  opposite  them,  respectively,  by  a,  b,  c 

Case  I. 

(53.)  Given  two  angles  and  a  side,  to  find  the  third  angle 
and  the  other  two  sides. 

To  find  the  third  angle,  add  the  given  angles  together,  and 
subtract  their  sum  from  180°. 

The  required  sides  may  be  found  by  Theorem  I.  The  pro- 
portion will  be, 

The  sine  of  the  angle  opposite  the  given  side :  the  given  side 

: :  the  sine  of  the  angle  opposite  the  required  side  :  the  re* 
quired  side. 

Ex.  1.  In  the  triangle  ABC,  there  are 
given  the  angle  A,  57°  15',  the  angle  B, 
35°  30',  and  the  side  c,  364,  to  find  the 
other  parts. 

The  sum  of  the  given  angles,  subtracted 


Plane    Trigonometry.  39 

from  180°,  leaves  87°  15'  for  the  angle  C.     Then,  to  find  the 
side  <z,  we  say,         sin.  C  :  c  : :  sin.  A.  :  a. 
By  natural  numbers, 

.998848  :  364  :  :  .841039  :  506.49=a. 
This  proportion  is  most  easily  worked  by  logarithms,  thus . 
As  the  sine  of  the  angle  C,  87°  15',  comp.,  0.000500 
Is  to  the  side  c^64,  2.561101 

So  is  the  sine  of  the  angle  A,  57°  15',       9.924816 
To  the  side  a,  306.49,  2.486417. 

To  find  the  side  b  : 

sin.  0  ',  c  :-,  sin.  B  :  ^. 
By  natural  numbers, 

.998848  :  364  : :  .580703  :  211.62=b. 
The  work  by  logarithms  is  as  follows  : 

sin.  C,  87°  15',  comp.,  0.000500 

:  c,  364,  2.561101 

: :  sin.  B,  35°  30',  9.763954 

:  b,  211.62,  2.325555. 

Ex.  2.  In  the  triangle  ABC,  there  are  given  the  angle  A, 
49°  25',  the  angle  C,-  63°  48',  and  the  side  c,  275,  to  find  the 
other  parts.        Ans.,  B=66°  47' ;  ^=232.766  ;  ^>=281.67. 

"7  Case  II. 

(54.)  Given  two  sides  and  an  angle  opposite  one  of  them, 
to  find  the  third  side  and  the  remaining  angles. 

One  of  the  required  angles  is  found  by  Theorem  I.  The 
proportion  is. 

The  side  opposite  the  given  angle  :  the  sine  of  that  angle 

:  :  the  other  given  side  :  the  sine  of  the  opposite  angle. 

The  third  angle  is  found  by  subtracting  the  sum  of  the  other 
two  from  180°  ;  and  the  third  side  is  found  as  in  Case  T. 

If  the  side  BC,  opposite  the  given  an-  ^ 

gle  A,  is  shorter  than  the  other  given  side 
AC,  the  solution  will  be  ambiguous  ;  that 

is,  two  different  triangles,  ABC,  AB'C,     y^.^  / \  ^ 

may  be  formed,  each  of  which  will  satisfy  a 

the  conditions  of  the  problem. 

^    The  numerical  result  is  also  ambiguous,  for  the  fourth  terra 


^ 


40 


Trig.*nometry. 


But  this  may  be 
C 


of  the  first  proportion  is  a  sine  of  an  angle, 
the  sine  either  of  the  acute  angle  AB'C,  or 
of  its  supplement,  the  obtuse  angle  ABC 
{Art.  27).  In  practice,  however,  there  will 
generally  be  some  circumstance  to  determ- 
ine whether  the  required  angle  is  acute  or  ^  B 
obtuse.  If  the  given  angle  is  obtuse,  Jjhere  can  be  no  ambi 
guity  in  the  solution,  for  then  the  remaining  angles  must  of 
course  be  acute. 

Ex.  1.  In  a  triangle,  ABC,  there  are  given  AC,  458,  BC, 
307,  and  the  angle  A,  28°  45',  to  find  the  other  parts. 


To  find  the  angle  B  : 


:  AC  :  sin.  B. 


BC  :  sin.  A 
By  natural  numbers, 

307  :  .480989  : :  458  :  .717566,  sin.  B,  the  arc  correspond- 
ing to  which  is  45°  51'  14",  or  134°  8'  46". 

This  proportion  is  most  easily  worked  by  logarithms,  thus 
BC,  307,  comp.,  7.512862 

:  sin.  A,  28°  45',  9.682135 

: :  AC,  458,  2.660865 

:  sin.  B,  45°  51'  14",  or  134°  8'  46",  9.855862. 
The  angle  ABC  is  134°  8'  46",  and  the  angle  AB'C,  45°  51' 
14".     Hence  the  angle  ACB  is  17°  6'  14",  and  the  angle  ACB'. 
105°  23' 46". 
To  find  the  side  AB  : 

sin.  A  :  CB  :  :  sin.  ACB 
By  logarithms, 

sin.  A,  28°  45',  comp., 
CB,  307, 
:  sin.  ACB,  17°  6'  14", 

AB,  187.72, 


:  AB. 

0.317865 

2.487138 
9.468502 
2.273505. 


To  find  the  side  AB' : 

sin.  A  :  CB'  :  :  sin.  ACB  :  AB'. 
By  logarithms, 

sin.  A,  28°  45',  comp.,  0.317865 

CB',  307,  2.487138 

:  sin.  ACB',  105°  23'  46",  9.984128 

AB',  615.36,  2.789131. 


Plane    Trigonometry.  4J 

Ex.  2.  In  a  triangle,  ABC,  there  are  given  AB,  532,  BC, 
358,  and  the  angle  C,  107°  40',  to  find  the  other  parts. 

Ans.  A=39°  52'  52"  ;  B=32°  27'  8"  ;  AC=299.6. 

In  this  example  there  is  no  ambiguity,  because  the  giveL 
angle  is  obtuse. 

Case  III. 

(55.)  Given  two  sides  and  the  included  angle,  to  find  the 
third  side  and  the  remaining  angles. 

The  sum  of  the  required  angles  is  foijnd  by  subtracting  the 
given  angle  from  180°.  The  difference  of  the  required  angles 
is  then  found  by  Theorem  II.  Half  the  difference  added  to 
half  the  sum  gives  the  greater  angle,  and,  subtracted,  gives 
the  less  angle.     The  third  side  is  then  found  by  Theorem  I. 

Ex.  1.  In  the  triangle  ABC,  the  angle  A  is  given  53°  8' ; 
the  side  c,  420,  and  the  side  b,  535,  to  find  the  remaining  parts. 

The  sum  of  the  angles  B  +  C=180°-53°  8' =126°  52'. 
Half  their  sum  is  63°  26'. 

Then,  by  Theorem  II.,   ' 

535+420  :  535-420  :  :  tang.  63°  26'  :  tang.  13°  32  25", 
which  is  half  the  difference  of  the  two  required  angles. 

Hence  the  angle  B  is  76°  58'  25",  and  the  angle  C,  49° 
53  35". 

To  find  the  side  a\ 

sin.  C  :  c  :  :  sin.  A  :  a=439.32. 

Ex.  2.  Griven  the  side  c,  176,  a,  133,  and  the  included  angle 
B,  73°,  to  find  the  remaining  parts. 

Ans.,  ^»=187.022,  the  angle  C,  64°  9'  3",  and  A,  42°  50'  57". 

Case  IV. 

(5Q.)  Given  the  three  sides,  to  find  the  angles. 

Let  fall  a  perpendicular  upon  the  longest  side  from  the  op- 
posite angle,  dividing  the  given  triangle  into  two  right-angled 
triangles.  The  two  segments  of  the  base  may  be  found  by 
Theorem  III.  There  will  then  be  given  the  hypothenuse  and 
one  side  of  a  right-angled  triangle  to  find  the  angl  is. 

Ex.  1.  In  the  triangle  ABC,  the  side  a  is  261,  the  side  b, 
345,  and  c,  395.     What  are  the  angles  ? 

Let  fall  the  perpendicular  CD  upon  AB. 


42  Trigonometry.  • 

Then,  by  Theorem  III., 

AB  :  AC  +  CB  :  :  AC-CB  :  AD-  DB ; 
or  395  :  606  : :  84  :  128.87. 

Half  the  difference  of  the  segments  added  to  half  their  sum 
gives  the  greater  segment,  and  subtracted  gives  the  less  seg- 
ment. .  JP 

Therefore,  AD  is  261.935,   and  BD,  / 

133.065.  V 

Then,  in  each  of  the  right-angled  tri-      / 
angles,  ACD,  BCD,  ,we  have  given  the    A  d  B 

hypothenuse  and  base,  to  find  the  angles  by  Case  II.  of  right- 
angled  triangles.     Hence 

AC  :  R  : :  AD  :  cos.  A=40°  36'  13" ; 
BC  :  R  : :  BD  :  cos.  B=59°  20'  52". 
Therefore  the  angle  C=80°  2'  ^^" . 

Ex.  2.  If  the  three  sides  of  a  triangle  are  150,  140,  and  130, 
what  are  the  angles  ? 

Ans.,  67°  22'  48",  59°  29'  23",  and  53"^  7'  49" 

Examples  for  Practice. 

1.  Griven  two  sides  of  a  triangle,  478  and  507,  and  the  in 
eluded  angle,  47°  30',  to  find  the  remaining  parts. 

2.  Griven  the  angle  A,  5Q°  34',  the  opposite  side,  a,  735,  and 
the  side  Z>,  576,  to  find  the  remaining  parts. 

3.  Given  the  angle  A,  Qi^""  40',  the  angle  B,  74°  20',  and  the 
side  <2,  275,  to  find  the  remaining  parts. 

4.  Given  the  three  sides,  742,  657,  and  379,  to  find  the  an- 
gles. 

5.  Given  the  angle  A,  116°  32',  the  opposite  side,  a,  492, 
and  the  side  c,  295,  to  find  the  remaining  parts. 

6.  Given  the  angle  C,  5Q>°  18',  the  opposite  side,  c,  184,  and 
the  side  Z>,  219,  to  find  the  remaining  parts. 

This  problem  admits  of  two  answers. 

INSTRUMENTS  USED  IN  DRAWING. 

(57.)  The  following  are  some  of  the  most  important  instru- 
ments used  in  drawing. 

I.  The  dividers  consist  of  two  legs,  revolving  upon  a  pivot 
at  (\x{Q  extremity.     The  joints  should  be  composed  of  two  dif- 


Plane    Trigonometr v. 


43 


ferent  metals,  of  unequal  hardness  :  one  part,  for  example,  of 
steel,  and  the  other  of 
brass  or  silver,  in  order 
that  they  may  move  upon 
e^h  other  with  greater 
freedom.  The  points  should  be  of  tempered  steel,  and  when 
the  dividers  are  closed,  they  should  meet  with  great  exactness. 
The  dividers  are  often  furnished  with  various  appendages, 
which  are  exceedingly  convenient  in  drawing.  Sometimes  one 
of  the  legs  is  furnished  with  an  adjusting  screw,  by  which  a 
slow  motion  may  be  given  to  one  of  the  points,  in  which  case 
they  are  called  hair  compasses.  It  is  also  useful  to  have  a 
movable  leg,  which  may  be  removed  at  pleasure,  and  other 
parts  fitted  to  its  place ;  as,  for  example,  a  long  beam  for 
drawing  large  circles,  a  pencil  point  for  drawing  circles  with 
a  pencil,  an  ink  point  for  drawing  black  circles,  &c. 

(58.)  II.  The  parallel  rule  consists  of  two  flat  rules,  made 
of  wood  or  ivory,  and  connected  together  by  two  cross-bars  of 


equal  length,  and  parallel  to  each  other.  This  instrument  iu 
useful  for  drawing  a  line  parallel  to  a  given  line,  through  a 
given  point.  For  this  purpose,  place  the  edge  of  one  of  the 
flat  rules  against  the  given  line,  and  move  the  other  rule  until 
'ts  edge  coincides  with  the  given  point.  A  line  drawn  along 
its  edge  will  be  parallel  to  the  given  line. 

(59.)  III.  The  protractor  is  used  to  lay  down  or  to  measure 

angles.    It  consists  of  a  sem-  

icircle,  usually  of  brass,  and 
is  divided  into  degrees,  and 
sometimes  smaller  portions, 
the  center  of  the  circle  be- 
ing indicated  by  a  small 
notch. 

To  lay  down  an  angle  with  the  protractor,  draw  a  base  line, 
and  apply  to  it  the  edge  of  the  protractor,  so  that  its  center 
shall  fall  at  the  angular  point      Count  the  degrees  contained 


14 


Trigonometry. 


■ 

m 

in  the  proposed  angle  on  the  limb  of  the  circle,  and  mark  the 
extremity  of  the  arc  with  a  fine  dot.  Remove  the  instrument, 
and  through  the  dot  draw  a  line  from  the  angular  point ;  it 
will  give  the  angle  required.  In  a  similar  manner,  the  in- 
clination of  any  two  lineg  may  be  measured  with  the  jfto- 
tractor. 

(60.)  IV.  The  plane  scale  is  a  ruler,  frequently  two  feet  in 
length,  containing  a  line  of  equal  parts,  chords,  sines,  tan- 
gents, &c.  For  a  scale  of  equal  parts,  a  line  is  divided  into 
inches  and  tenths  of  an  inch,  or  half  inches  and  twentieths. 
When  smaller  fractions  are  required,  they  are  obtained  by 
means  of  the  diagonal  scale,  which  is  constructed  in  the  fol- 
lowing manner.     Describe  a  square  inch,  ABCD,  and  divide 

4  -.3 2 1_ A.2  4.6.g    B 

'  '  02 
04 
06 
.08 

DE  'C 

each  of  its  sides  ihto  ten  equal  parts.  Draw  diagonal  lines 
from  the  first  point  of  division  on  the  upper  line,  to  the  second 
on  the  lower  ;  from  the  second  on  the  upper  line,  to  the  third 
on  the  lower,  and  so  on.  Draw,  also,  other  lines  parallel  i: 
AB,  through  the  points  of  division  of  BC.  Then,  in  the  trian- 
gle ADE,  the  base,  DB,  is  one  tenth  of  an  inch;  and,  since 
the  line  AD  is  divided  into  ten  equal  parts,  and  through  the 
points  of  division  lines  are  drawn  parallel  to  the  base,  forming 
nine  smaller  triangles,  the  base  of  the  least  is  one  tenth  of  DE, 
that  is,  .01  of  an  inch;  the  base  of  the  second  is  .02  of  an  inch ; 
the  third,  .03,  and  so  on.  Thus  the  diagonal  scale  furnishes 
us  hundredths  of  an  inch.  To  take  off"  from  the  scale  a  line 
of  given  length,  as,  for  example,  4.45  inches,  place  one  foot  of 
the  dividers  at  F,  on  the  sixth  horizontal  line,  and  extend  tho 
other  foot  to  G-,  the  fifth  diagonal  line. 

A  half  inch  or  less  is  frequently  subdivided  in  the  same 
manner. 

(61.)  A  line  of  chords,  commonly  marked  cho.,  is  found  on 
most  plane  scales,  and  is  useful  in  setting  off  angles.  To  form 
this  line,  describe  a  circle  with  any  convenient  radius,  and  di- 
vide the  circumference  into  degrees.     Let  the  length  of  the 


Plane   Trigonometry^.  45 

chords  foi  every  degree  of  the  quadrant  be'deter mined  and  laid 
off  on  a  scale  :  this  is  called  a  line  of  chords. 

Since  the  chord  of  60°  is  equal  to  radius,  in  order  to  lay 


Chords 

10 

20 

So        40        SO       6o 

no     80  -.  90 

Sines 

10 

20 

SO      4p     50    foTOSJo 

30      40          So  Secants 

60 

Tang. 

^ 

20 

50               40 

50                         eo 

down  an  angle,  we  take  from  the  scale  the  chord  of  60"",  and 
with  this  radius  describe  an  arc  of  a  circle.  Then  take  from 
the  scale  the  chord  of  the  given  angle,  and  set  it  off  upon  the 
former  arc.  Through  these  two  points  of  division  draw  lines 
to  the  center  of  the  circle,  and  they  will  contain  the  required 
angle. 

The  line  of  sines,  commonly  marked  sin.,  exhibits  the  lengths 
of  the  sines  to  every  degree  of  the  quadrant,  to  the  same  ra- 
dius as  the  line  of  chords.  The  line  of  tangents  and  the  line 
of  secants  are  constructed  in  the  samiB  manner.  Since  the  sine 
of  90°  is  equal  to  radius,  and  the  secant  of  0°  is  the  same,  the 
graduation  on  the  line  of  secants  begins  where  the  line  of  sine3 
ends. 

On  the  back  side  of  the  plane  scale  are  often  found  lines  rep- 
resenting the  logarithms  of  numbers,  sines,  tangents,  &c.  This 
is  called  Grunter'3  Scale. 

(62.)  Y.  The  Sector  is  a  very  convenient  instrument  in 
drawing.  It  consists  of 
two  equal  arms,  mova- 
ble about  a  pivot  as  a 
center,  having  several 
scales  drawn  on  the 
faces,  some  single,  oth- 
ers double.  The  single  scales  are  like  those  upon  a  common 
Grunter's  scale.  The  double  scales  are  those  which  proceed 
from  the  center,  each  being  laid  twice  on  the  same  face  of  the 
instrument,  viz.,  once  on  each  leg.  The  double  scales  are  a 
scale  of  lines,  marked  Lin.  or  L. ;  the  scale  of  chords,  sines, 
&c.  On  each  arm  of  the  sector  there  is  a  diagonal  line,  which 
diverges  from  the  central  point  like  the  radius  of  a  circle,  and 
these  diagonal  lines  are  divided  into  equal  parts. 

The  advantage  of  the  sector  is  to  enable  us  to  draw  a  line 


46  Trigonometry. 

upon  paper  to  any  scale ;  as,  for  example,  a  scale  of  6  feet  to 
the  inch.  For  this  purpose,  take  an  inch  with  the  dividers 
from  the  scale  of  inches  ;  then,  placing  one  foot  of  the  dividers 
at  6  on  one  arm  of  the  sector,  open  the  sector  until  the  other 
foot  reaches  to  the  same  numher  on  the  other  arm.  Now,  re- 
garding the  lines  on  the 
sector  as  the  sides  of  a 
triangle,  of  which  the 
line  measured  from  6  on 
one  arm  to  6  on  the  oth- 
er arm  is  the  base,  it  is 
plain  that  if  any  other  line  be  measured  across  the  angle  of 
the  sector,  the  bases  of  the  triangles  thus  formed  will  be  pro- 
portional to  their  sides.  Therefore,  a  line  of  7  feet  will  be  rep- 
resented by  the  distance  from  7  to  7,  and  so  on  for  other  lines. 

The  sector  also  contains  a  line  of  chords,  arranged  like  the 
line  of  equal  parts  already  mentioned.  Two  lines  of  chords 
are  drawn,  one  on  each  arm  of  the  sector,  diverging  from  the 
central  point.  This  double  line  of  chords  is  more  convenient 
than  the  single  one  upon  the  plane  scale,  because  it  furnishes 
chords  to  any  radius.  If  it  be  required  to  lay  down  any  angle, 
as,  for  example,  an  angle  of  25°,  describe  a  circle  with  any 
convenient  radius.  Open  the  sector  so  that  the  distance  from 
(30  to  60,  on  the  line  of  chords,  shall  be  equal  to  this  radius. 
Then,  preserving  the  same  opening  of  the  sector,  place  one  foot 
of  the  dividers  upon  the  division  25  on  one  scale,  and  extend 
the  other  foot  to  the  same  number  upon  the  other  scale :  this 
distance  will  be  the  chord  of  25  degrees,  which  must  be  set  off 
upon  the  circle  first  described. 

The  lines  of  sines,  tangents,  &c.,  are  arranged  in  the  same 
manner. 

(63.)  By  means  of  the  instruments  now  enumerated,  all  the 
cases  in  Plane  Trigonometry  may  be  solved  mechanically. 
The  sides  and  angles  which  are  given  are  laid  down  accord- 
ing to  the  preceding  directions,  and  the  required  parts  are  then 
measured  from  the  same  scale.  The  student  will  do  well  to 
exercise  himself  upon  the  following  problems  : 

1.  Given  the  angles  and  one  side  of  a  triangle,  to  find,  by 
construction,  the  other  two  sides. 


Plane    Trigonometry.  47 

Draw  an  indefinite  straight  line,  and  from  the  scale  of  equal 
parts  lay  off  a  portion,  AB,  equal  to  the  given  side.  From 
each  extremity  lay  off  an  angle  equal  to  one  of  the  adjacent  an- 
gles, hy  means  of  a  protractor  or  a  scale  of  chords.  Extend 
the  two  lines  till  they  intersect,  and  measure  their  lengths  upon 
the  same  scale  of  equal  parts  which  was 
used  in  laying  off  the  base. 

Ex.  1.  Griven  the  angle  A,  45°  30',  the 
angle  B,  35°  20',  and  the  side  AB,  43*^ 
rods,  to  construct  the  triangle,  and  find 
the  lengths  of  the  sides  AC  and  BC. 

The  triangle  ABC  may  be  constructed  of  any  dimensions 
whatever ;  all  which  is  essential  is  that  its  angles  be  made 
equal  to  the  given  angles.  We  may  construct  the  triangle 
upon  a  scale  of  100  rods  to  an  inch,  in  which  case  the  side  AB 
will  be  represented  by  4.32  inches ;  or  we  may  construct  it 
upon  a  scale  of  200  rods  to  an  inch ;  that  is,  100  rods  to  a  half 
inch,  which  is  very  conveniently  done  from  a  scale  on  which 
a  half  inch  is  divided  like  that  described  in  Art.  60 ;  or  we 
may  use  any  other  scale  at  pleasure.  It  should,  however,  be 
remembered,  that  the  required  sides  must  be  measured  upon 
the  same  scale  as  the  given  sides. 

Ex.  2.  G-iven  the  angle  A,  48°,  the  angle  C,  113°,  and  the 
side  AC,  795,  to  construct  the  triangle. 

II.  Given  two  sides  and  an  angle  opposite  one  of  them,  to 
find  the  other  parts. 

Draw  the  side  which  is  adjacent  to  the  given  angle.  From 
one  end  of  it  lay  off  the  given  angle,  and  extend  a  line  indefin- 
itely for  the  required  side.  From  the  other  extremity  of  the 
first  side,  with  the  remaining  given  side  for  ladius,  describe 
an  arc  cutting  the  indefinite  line.  The  point  of  intersection 
will  determine  the  third  angle  of  the  triangle. 

Ex.  1.  Given  the  angle  A,  74°  45',  the  side  AC,  432,  and 
the  side  BC,  475,  to  construct  the  triangle,  and  find  the  other 
parts. 

Ex.  2.  G-iven  the  angle  A,  105°,  the  side  BC,  498,  and  the 
side  AC,  375,  to  construct  the  triangle. 

III.  Given  two  sides  and  the  included  angle,  to  find  the 
other  parts. 


48  Trigonometry. 

Draw  one  of  the  given  sides.  From  one  end  of  it  lay  off  the 
given  angle,  and  draw  the  other  given  side,  making  the  re- 
quired angle  with  the  first  side.  Then  connect  the  extremities 
of  the  two  sides,  and  there  will  be  formed  the  triangle  required. 

Ex.  1.  G-iven  the  angle  A,  37°  25',  the  side  AC,  675,  and 
the  side  AB,  417,  to  construct  the  triangle,  and  find  the  other 
parts. 

Ex.  2.  G-iven  the  angle  A,  75°,  the  side  AC,  543,  and  the 
side  AB,  721,  to  construct  the  triangle. 

lY.  Given  the  three  sides,  to  find  the  angles. 

Draw  one  of  the  sides  as  a  base ;  and  from  one  extremity 
of  the  base,  with  a  radius  equal  to  the  second  side,  describe 
an  arc  of  a  circle.  From  the  other  end  of  the  base,  with  a 
radius  equal  to  the  third  side,  describe  a  second  arc  intersect- 
ing the  former ;  the  point  of  intersection  will  be  the  third  an- 
gle of  the  triangle. 

Ex.  1.  Given  AB,  678,  AC,  598,  and  BC,  435,  to  find  the 
angles. 

Ex.  2.  G-iven  the  three  sides  476,  287,  and  354,  to  find  the 

angles. 

• 

Values  of  the  Sines,  Cosines,  Sfc,  of  certain  Angles 
(64.)  We  propose  now  to  examine  the  changes  which  the 
tines,  cosines,  &c.,  undergo  in  the  dif- 
ferent quadrants  of  a  circle.  Draw 
two  diameters,  AB,  DE,  perpendicu- 
lar to  each  other,  and  suppose  one  of 
them  to  occupy  a  horizontal  position, 
the  other  a  vertical.  The  angle  ACD 
is  called  the  first  quadrant,  the  angle 
DCB  the  second  quadrant,  the  angle 
BCE  the  third  quadrant,  and  the  an- 
gle ECA  ^Q  fourth  quadrant;  that  is,  the  first  quadrant  is 
above  the  horizontal  diameter,  and  on  the  right  of  the  vertical 
diameter ;  the  second  quadrant  is  above  the  horizontal  diame- 
ter, and  on  the  left  of  the  vertical,  and  so  on. 

Suppose  one  extremity  of  the  arc  remains  fixed  in  A,  while 
the  other  extremity,  marked  F,  runs  round  the  entire  circum- 
ference in  the  direction  ADBE. 


Plane    Trigonometry.  49 

Wbin  the  point  F  is  at  A,  or  when  the  arc  AF  is  zero,  the 
sine  is  zero.  As  the  point  F  advances  toward  D,  the  sine  in- 
creases ;  and  when  the  arc  AF  becomes  45°,  the  triangle  CFG 
being  isosceles,  we  have  FGr  :  OF  : :  1  :  v/2  {Geom.^  Prop.  11, 
Cor.  3,  B.  IV.) ;  or    sin.  45°  :  R  : :  1  :  ^/2. 

Hence,  sin.  45°=— ^=iRx/2. 

The  sine  of  30°  is  equal  to  half  radius  (Art.  22).  Also,  since 
sin.  A=  VR'^— COS.  ''A,  the  sine  of  60°,  which  is  equal  to  the  co- 
sine of  30%  =  yfW^^'=  VJR^zriR  V3. 

The  arc  AF  continuing  to  increase,  the  sine  also  increaseis 
till  F  arrives  at  D,  at  which  point  the  sine  is  equal  to  the  ra- 
dius ;  that  is,  the  sine  of  90° =R. 

As  the  point  F  advances  from  D  toward  B,  the  sines  dimin- 
ish, and  become  zero  at  B  ;  that  is,  the  sine  of  180° =0. 

In  the  third  quadrant,  the  sine  increases  again,  becomes 
oqual  to  radius  at  E,  and  is  reduced  to  zero  at  A. 

(^5.)  When  the  point  F  is  at  A,  the  cosine  is  equal  to  ra- 
dius. As  the  point  F  advances  toward  D,  the  cosine  decreases, 
and  the  cosine 'of  45°= sine  45°=|R\/2.  The  arc  continuing 
to  increase,  the  cosine  diminishes  till  F  arrives  at  D,  at  which 
point  the  cosine  becomes  equal  to  zero.  The  cosine  in  the  sec- 
ond quadrant  increases,  and  becomes  equal  to  radius  at  B ;  in 
the  third  quadrant  it  decreases,  and  becomes  zero  at  E  ;  in  the 
fourth  ^quadrant  it  increases  again,  and  becomes  equal  to  ra- 
dius at  A. 

(foQ.)  The  tangent  begins  with  zero  at  A,  increases  with  the 
arc,  and  at  45°  becomes  equal  to  radius.  As  the  point  F  ap- 
proaches D,  the  tangent  increases  very  rapidly ;  and  when  the 
difference  between  the  arc  and  90°  is  less  than  any  assignable 
quantity,  the  tangent  is  greater  than  any  assignable  quantity. 
Hence  the  tangent  of  90°  is  said  to  be  infinite. 

In  the  second  quadrant  the  tangent  is  at  first  infinitely  great, 
and  rapidly  diminishes  till  at  B  it  is  reduced  to  zero.  In  the 
third  quadrant  it  increases  again,  becomes  infinite  at  E,  and  is 
reduced  to  zero  at  A. 

The  cotangent  is  equal  to  zero  at  D  and  E,  and  is  infinite  at 
A  and  B. 

(67.)  The  secant  begins  with  radius  at  A,  increases  through 

D 


50  Trigonometry. 

the  first  quadrant,  and  becomes  infinite  at  D  ;  diminishes  in 

the  second  quadrant,  till  at  B  it  is 

equal  to  the  radius ;  increases  again 

in  the  third  quadrant,  and  becomes 

infinite  at  E  ;  decreases  in  the  fourth 

quadrant,  and  becomes  equal  to  the 

radius  at  A. 

The  cosecant  is  equal  to  radius  at 
D  and  E,  and  is  infinite  at  A  and  B. 

(68.)  Let  us  now  consider  the  al- 
gebraic signs  by  which  these  lines  are  to  be  distinguished.  In 
the  first  and  second  quadrants,  the  sines  fall  above  the  diame- 
ter AB,  while  in  the  third  and  fourth  quadrants  they  fall  be- 
low. This  opposition  of  directions  ought  to  be  distinguished 
by  the  algebraic  signs ;  and  if  one  of  these  directions  is  re- 
garded as  positive,  the  other  ought  to  be  considered  as  nega- 
tive. It  is  generally  agreed  to  consider  those  sines  which  fall 
above  the  horizontal  diameter  as  positive  ;  consequently,  those 
which  fall  below  must  be  regarded  as  negative.  That  is,  the 
sines  are  positive  in  the  first  and  second  quadrants,  and  nega- 
tive in  the  third  and  fourth. 

In  the  first  quadrant  the  cosine  falls  on  the  right  of  DE, 
but  in  the  second  quadrant  it  falls  on  the  left.  These  two  lines 
should  obviously  have  opposite  signs,  and  it  is  generally  agreed 
to  consider  those  which  fall  to  the  right  of  the  vertical  diam- 
eter as  positive  ;  consequently,  those  which  fall  to  the  left  must 
be  considered  negative.  That  is,  the  cosines  are  positive  in 
the  first  and  fourth  quadrants,  and  negative  in  the  second  and 
third. 

(69.)  The  signs  of  the  tangents  are  derived  from  those  of 

the  sines  and  cosines.     For  tanpr.  =— -*  (Art.  28).     Hence, 

°  COS.     ^  ' 

when  the  sine  and  cosine  have  like  algebraic  signs,  the  tan- 
gent will  be  positive ;.  when  unlike,  negative.    That  is,  the  tan 
gent  is  positive  in  the  first  and  third  quadrants,  and  negative 
in  the  second  and  fourth. 

T>2 

Also,  cotangent  =- (Art.  28) ;  hence  the  tangent  and 

<v)tangent  have  always  the  same  sign. 


.Plane   Trigonometry.  51 

"We  have  seen  that  sec.  = ;  hence  the  secant  must  have 

COS. 

the  same  sign  as  the  cosine. 

Also,  cosec.  =-. — ;  hence  the  cosecant  must  have  the  same 
'  sm. 

sign  as  the  sine. 

(70.)  The  preceding  results  are  exhibited  in  the  following 

tables,  which  should  be  made  perfectly  familiar : 

First  quad.    Second  quad.    Third  quad.    Fourth  quad. 

Sine  and- cosecant,  +  +  —  — 

Cosine  and  secant,  4-  _  _  _^ 

Tangent  and  cotangent,     +  —  +  - 


0° 

90° 

180° 

270° 

360° 

Sine, 

0 

+R 

0 

-R 

0 

Cosine, 

+11 

0 

-R 

0 

+R 

Tangent, 

0 

00 

0 

00 

0 

Cotangent, 

00 

0 

GO 

0 

00 

Secant, 

+R 

00 

-R 

GO 

+R 

Cosecant, 

00 

+R 

00 

-R 

GO 

(71.)  In  Astronomy  we  frequently  have  occasion  to  consider 
arcs  greater  than  360°.  But  if  an  entire  circumference,  or  any 
number  of  circumferences,  be  added  to  any  arc,  it  will  termin- 
ate in  the  same  point  as  before.  Hence,  if  C  represent  an  en- 
tire circumference,  or  360°,  and  A  any  arc  whatever,  we  shall 
have 

sin.  A=sin.  (C+A)=sin.  (2C+A)=sin.  (3C+A)=,  &c. 

The  same  is  true  of  the  cosine,  tangent,  &c. 

We  generally  consider  those  arcs  as  positive  which  are  esti- 
mated from  A  in  the  direction  ADBE.  If,  then,  an  arc  were 
estimated  in  the  direction  AEBD,  it  should  be  considered  as 
negative ;  that  is,  if  the  arc  AF  be  considered  positive,  AH 
must  be  considered  negative.  But  the  latter  belongs  to  the 
fourth  quadrant ;  hence  its  sine  is  negative.  Therefore,  sin. 
(— A)=— sin.  A. 

The  cosine  CGr  is  the  same  for  both  the  arcs  AF  and  AH. 
Hence,  cos.  (— A)=cos.  A. 

Also,  tang.  ( — A) = —tang.  A. 

And  cot.  ( — A)  = — cot.  A. 


52 


Trigonometry. 


TRIGONOMETRICAL  FORMULAE. 

(72.)  Expressions  for  the  sine  and  cosine  of  the  sum  and 
difference  of  two  arcs. 

Let  AB  and  BD  represent  any  two  given  arcs ;  take  BE 
equal  to  BD :  it  is  required  to  find  an 
expression  for  the  sine  of  AD,  the  sum, 
and  of  AE,  the  difference  of  these  arcs. 

Put  AB=a,  and  BD=Z>;  then  AD= 
a-\-b^  and  AE =a—b.  Draw  the  chord 
DE,  and  the  radius  CB,  which  may  be 
represented  by  R.  Since  DB  is  by 
construction  equal  to  BE,  DF  is  equal 
to  FE,  and  therefore  DE  is  perpendic-  ^  Kl  IK  G-  A 
ular  to  CB.  Let  fall  the  perpendiculars  EG-,  BH,  FI,  and  DK 
upon  AC,  and  draw  EL,  FM  parallel  to  AC. 

Because  the  triangles  BCH,  FCI  are  similar,  we  have 
CB  :  CF  : :  BH  ;  FI :  or  R  :  cos.  b  : :  sin.  a  :  FL 


Therefore, 

Also,  CB 
Therefore, 


CF 


FI= 
:  CH  :  CI 


sin.  a  COS.  b 


CI= 


R 
;  or  R  :  cos.  h 
COS.  a  COS.  b 


:  COS.  a 


CI. 


R 


The  triangles  DFM,  CBH,  having  their  sides  perpendicular 
each  to  each,  are  similar,  and  give  the  proportions 

CB  :  DF  : :  CH  :  DM ;  or  R  :  sin.  b  : :  cos.  a  :  DM. 
cos.  a  sin.  b 


Hence 

Also,  CB 
Hence 


DM= 


R 


DF  : :  BH  :  FM :  or  R 


sm. 


sm.  a 


FM. 


FM= 


sin.  a  sin.  b 


R 


But  FI+DM=DK  =:sin.  (a+b) ; 

and  CI-FM=CK  =cos.  (a+b). 

Also,  FI-FL  =EG-=sin.  (a-b) ; 

and  CI+EL  =  Ca=cos.  (a-b). 

__  .     ,       TV     sin.  a  COS.  b+cos.  a  sin.  b 

Hence,  sm.  (a+b)= =p — 

COS.  a  COS.  6— sin.  a  sin.  b 


COS.  {a+b)= 


R 


(1) 
(2) 


Plane   Trigonometry.  53 

.     .       _.     sin.  a  COS.  6— cos.  a  sin.  b  ,«. 
Bm,(a-'b)= g^ (3) 

.       ,.     COS.  o  COS.  6+ sin.  a  sin.  Z>  ,,^ 
COS.  (a-b)= g (4) 

(73.)  Expressions  for  the  sine  and  cosine  of  a  double  arc. 

If,  in  the  formulas  of  the  preceding  article,  we  make  6=o 

the  first  and  second  will  become 

.     rt      2  sin.  a  cos.  a 
sm.  2a= ^ , 

COS.  ^a— sin.  ^a 
cos.  2a= ^ — . 

Making  radius  equal  to  unity,  and  substituting  the  values  ol 
sin.  a,  COS.  a,  &c.,  from  Art.  28,  we  obtain 

.  2  tang,  a 

sm.  2a=^  ,  X        a  > 
1+tang.  a 

1-tang.  'a 

COS.  2a=r7-—: — ^—r- 

1+tang.  ^a 

(74.)  Expressions  for  the  sine  and  cosine  of  half  a  given 
arc. 

Tf  we  put  ^a  for  a  in  the  preceding  equations,  we  obtain 
2  sin.  ^a  cos.  ^a 


sm.  a= 


R 


cos.  ^ia— sm.  %a 

COS.  a= 15 ^—. 

-"' 

We  may  also  find  the  sine  and  cosine  of  ^a  in  terms  of  a. 

Since  the  sum  of  the  squares  of  the  sine  and  cosine  is  equal 

to  the  square  of  radius,  we  have 

cos.  'Ja+sin.  ''^a=R^ 

And,  from  the  preceding  equation, 

COS.  "^a— sin.  'Ja=R  cos.  a. 

If  we  subtract  one  of  these  from  the  other,  we  have 

2  sin.  ''Ja=R'— R  cos.  a. 

And,  adding  the  same  equations, 

2  COS.  'Ja=R''+R  COS.  a. 

Hence,  sin.  Ja=  V^R''— ^R  cos.  a; 

COS.  ^a=  V^R'+JR  cos.  a. 

(75.)  Expressions  for  the  products  of  sines  and  cosines. 

By  adding  and  subtracting  the  formulas  of  Art.  72,  we  obtain 


54  Trigonometry. 

2 
sin.  (a+b) +s'm.  (a— Z>)=—  sin.  a  cos.  b. 

2 
sin.  (a+^)— sin.  (a—b)=^  cos.  a  sin.  b  ; 

2 

COS.  (a+^)+cos.  (a—b)=^  COS.  a  cos.  ft; 

2 
cos.  (a—b)—cos,  ((i+b)=^  sin.  a  sin.  b. 

If,  in  these  formulas,  we  make  a+^=A,  and  a—b=B  ;  that 
is,  a=i(A+B),  and  6=J(A— B),  we  shall  have 

2 
sin.  A+sin.  B^g-  sin.  J(A+B)  cos.  i(A-B)     (1) 

2 

sin.  A-sin.  B=g-  sin.  i(A-B)  cos.  |(A+B)     (2) 

2 

cos.  A+cos.  B=p  cos.  i(A+B)  cos.  i(A-B)     (3) 

2 
COS.  B-cos.  A=p  sin.  i(A+B)  sin.  i(A-B)     (4) 

(76.)   Dividing  formula  (1)  by  (2),  and  considering  that 

sin.  a    tanff.  a  , ,       ^^. 

=—~—  (Art.  28),  we  have 

cos.  a        R       ^ 

sin.  A+sin.  Bsin.  |(A+B)  cos.  ^(A-B)_tang.  |(A+B)  ^ 

sin.  A-sin.  B~sin.  i(A-B)  cos.  i(A+B)~tang.  i(A-B) ' 

that  is, 

The  sum  of  the  sines  of  two  arcs  is  to  their  difference^  as 

the  tangent  of  half  the  sum  of  those  arcs  is  to  the  tangent  of 

half  their  difference. 

cos      Pot 
JJividing  formula  (3)  by  (4),  and  considering  that  -r-^=-t^ 

sin.      ix> 

=7 (Art.  28),  we  have 

cos.  A+cos.  Bcos.  K-^+^)  <^^3.  ^(A--B)_  cot.  i(A+B) 
COS.  B-cos.  A~sin.  i(A+B)  sin.  i(A-B)""tang.^(A-B)' 

that  is. 

The  sum  of  the  cosines  of  two  arcs  is  to  their  difference^  as 

the  cotangent  of  half  the  sum  of  those  arcs  is  to  the  tangent 

of  half  their  difference. 


Plane   Trigonometry.  55 

From  the  first  formula  of  Art.  74,  by  substituting  A+B  for 

t,  we  have 

8sin.^(A+B)Xcos.KA+B) 
sm.  (A+±)j  — ^5 . 

Dividing  formula  (1),  Art.  75,  by  this,  we  obtain 

sin.  A+sin.  B_sin.  i(A+B)  cos.  ^(A-B)_cos.  i(A-B)  ^ 

sin.  (A+B)  "sin.  i(A+B)  cos.  i(A+B)""cos.  i(A+B) ' 

that  is,  t, 

The  sum  of  the  sines  of  two  arcs  is  to  the  sine  of  their 

sum,  as  the  cosine  of  half  the  difference  of  those  arcs  is  to 

the  cosine  of  half  their  sum. 

If  we  divide  equation  (1),  Art.  72,  by  equation  (3),  we  shall 

have 

sin.  [a+b)     sin.  a  cos.  6+cos.  a  sin.  b 

sin.  {a—b)     sin.  a  cos.  6— cos.  a  sin.  b' 

By  dividing  both  numerator  and  denominator  of  the  second 

11                      7        11.-      tang.  ^     sin.  _ 

member  by  cos.  a  cos.  £>,  and  substitutmg  -^5 —  for ,  we  ob- 

Xb  COS. 

sin.  ia-Vb)    tang.  «+tang.  ^     ,,    ,  . 

tain  -^, rT=r-^ — \     ^   .;  that  is, 

sm.  (a— o)    tang,  a— tang.  6 

The  sine  of  the  sum  of  two  arcs  is  to  the  sine  of  their  dif- 
ference, as  the  sum  of  the  tangents  of  those  arcs  is  to  the 
difference  of  the  tangents. 

From  equation  (3),  Art.  72,  by  dividing  each  member  by  cos 
a  COS.  b,  we  obtain 

sin.  [a—b)  _sin.  a  cos.  ^— cos.  a  sin.  Z>_tang.  a— tang,  b  ^ 

COS.  a  COS.  b~  R  cos.  a  cos.  b  R'  ' 

that  is, 

The  sine  of  the  difference  of  two  arcs  is  to  the  product  of 
their  cosines,  as  the  difference  of  their  tangents  is  to  the 
square  of  radius. 

(77.)  Expressions  for  the  tangents  of  arcs. 

If  we  take  the  expression  tang.  (a+b)= '       ,v     (Art. 

28),  and  substitute  for  sin.  (a+b)  and  cos.  (a+b)  their  values 

given  in  Art.  72,  we  shall  find 

-.     R  (sin.  a  cos.  b+cos.  a  sin.  b) 

tang.  {a+b)= — ^ r = = — j— - 

°  '        COS.  a  COS.  6— sm.  a  sm.  b 


56  Trigonometry. 

.           COS.  a  tang,  a       ,    .     ,     cos.  h  tang,  h 
But  sin  a=^ ^  ,  and  sin.  ^>= ^ — ^—  (Art.  28). 

If  we  substitute  these  values  in  the  preceding  equation,  and 

divide  all  the  terms  by  cos.  a  cos.  6,  we  shall  have 

,.     R'' (tansr.  a+tanff.  &) 
tang.  (a-^h)=  -p^    .  ^         ,     ^  ^. 
^  ^        '      R  —  tang,  a  tang,  h 

In  like  manner  we  shall  find 

R'  (tang,  a -tang,  h) 

•  tang.  (a-&)=  -pa',  ,  ° r — ^-r- 

°  ^         '      R  +tang.  a  tang.  6 

Suppose  6=«,  then 

2R'  tang,  a 

tang.  2a= 5-^—7 ^. 

°  R— tang,  a 

Suppose  ^=2a,  then 

R''  (tansr.  a + tang.  2a) 

tang.  3a=  -pa^   .  ° 7 —     o    • 

^  R  —  tang,  a  tang.  2a 

In  the  same  manner  we  find 

,,     cot.  a  cot.  6— R' 

cot.  (a+o)= r-^-^ — 7 , 

^        '       cot.  6+cot.  a  ' 

^       ,.     cot.  a  cot.  6+R' 

cot.  («—«>)= TT^ 7 . 

^        '       cot.  o— cot.  a 
(78.)  "When  the  three  sides  of  a  triangle  are  given,  the  an- 
gles may  be  found  by  the  formula 

sm.  iA=RV Y^ , 

where  S  represents  half  the  sum  of  the  sides  a,  b,  and  c. 


Demonstration. 

Let  ABC  be  any  triangle ;  then  (Geom.,  Prop.  12,  B.  IV.), 

BC'=AB»+AC'-2ABxAD.  ^ 

,^    AB'+AC^-BC^ 
Hence,      AD= ^^ . 

But  in  the  right-angled  triangle  ACD, 

we  have  "^ 

R  :  AC  : :  cos.  A  :  AD. 

*     RXAD 
Hence,  cos.  A= — 7-^; — ; 

AC     \ 

or,  by  substituting  the  value  of  AD, 

AB'+AC'-BC' 


cos.  A=Rx- 


2ABxAC 


Jt'LANE    Trigonometry.  57 

Let  a^  b^  c  represent  the  sides  opposite  the  angles  A,  B,  C ; 

then,  cos.  A=KX — ^a • 

By  Art.  74,  we  have  2  sin.  'JA=R'-R  cos.  A. 
Substituting  for  cos.  A  its  value  given  above,  we  obtain 

...•U=R-a-x5:^U.x?5£±^'. 

_Wx(a+b-c)(a+c-b) 

~  Wc  • 

Put  S=J(a+&+c),  and  we  obtain,  after  reduction, 


.^.^^=W^=^ 


\-c] 
be 
Tn  the  same  manner  we  find 


.in.iB=Rs/BS 


i=Rx/i? 


a)  {^-b) 


sm.  ^  ^     ^.  ^  , 

^  ^  ab 

Ex.  1.  What  are  the  angles  of  a  plane  triangle  whose  sides 
are  432,  543,  and  654? 

HereS=814.5;  8-6=382.5;  S-c=271.5. 

log.  382.5  2.582631 

log.  271.5  2.433770 

log.  ^>,  432  comp.    7.364516 

log.  c,  543  comp.    7.265200 

2)  19.646117 
sin.  JA,  41°  42'  36^".  9.823058. 

Angle  A=83°  25'  13". 
In  a  similar  manner  we  find  the  angle  B=41°  0'  39",  and 
the  angle  C=55°  34'  8". 

Ex.  2.  What  are  the  angles  of  a  plane  triangle  whose  sides 
are  245,  219,  and  91  ? 

(79.)  On  the  computation  of  a  table  of  sines ^  cosines^  Sfc. 
In  computing  a  table  of  sines  and  cosines,  we  begin  with 
finding  the  sine  and  cosine  of  one  minute^  and  thence  deduce 
the  sines  and  cosines  of  larger  arcs.  The  sine  of  so  small  an 
angle  as  one  minute  is  nearly  equal  to  the  corresponding  aio. 
The  radius  being  taken  as  unity,  the  semicircumference  is 


58  Trigonometry. 

known  to  be  3.14159.  This  being  divided  successively  by  180 
and  60,  gives  .0002908882  for  the  arc  of  one  minute,  which 
may  be  regarded  as  the  sine  of  one  minute. 

The  cosine  of  V=  VI- sin.'^^ 0.9999999577. 
The  sines  of  very^small  angles  are  nearly  proportional  to  the 
angles  themselves.  We  might  then  obtain  several  other  sines 
by  direct  proportion.  This  method  will  give  the  sines  correct 
to  five  decimal  places,  as  far  as  two  degrees.  By  the  follow- 
ing method  they  may  be  obtained  with  greater  accuracy  for 
t:he  entire  quadrant. 

By  Art.  75,  we  have,  by  transposition, 

sin.  (a+b)=2  sin.  a  cos.  6— sin.  (a—b), 

COS.  {a-{-b)=2  cos.  a  cos.  6— cos.  (a—b), 

[f  we  make  a=bj  2b,  Sb,  &c.,  successively,  we  shall  have 

sin.  2b=2  sin.  b  cos.  b  ; 

sin.Sb=2s'm.2bcos.b—8m.b, 

sin.  46=2  sin.  2b  cos.  6— sin.  26, 

&c.,  &c. 

COS.  26=2  COS.  b  COS.  6—1 ; 
COS.  36=2  COS.  26  cos.  6— cos.  6  ; 
COS.  46=2  cos.  36  cos,  6— cos.  26, 
&c.,  &c. 

Whence,  making  6=1',  we  have 

sin.  2'=2  sin.  V  cos.  1'  =.000582 ; 

sin.  3'=2  sin.  2'  cos.  I'-sin.  1'=.000873; 
sin.  4'=2  sin.  3'  cos.  I'-sin.  2'=.001164, 

&c.,  &c. 

COS.  2'=2  cos.  1'  COS.  V-        1  =0.999999  ; 

COS.  3'=2  COS.  2'  COS.  I'-cos.  l'=0.999999 ; 

COS.  4'=2  COS.  3'  COS.  I'-cos.  2'=0.999999, 

&c.,  &c. 

The  tangents,  cotangents,  secants,  and  cosecants  are  easily  . 

derived  from  the  sines  and  3osines.     Thus, 


sin.  1'  ^  .    cos.  1' 


tang.  1'= T7  cot.  1 


COS.  1'  V  sin.  1' 

1  .  1 


sec.  1'= r-; ;  cosec.  1 


COS.  1'  '         sin.  1' ' 

&c.,  &c 


>^  BOOK  III. 

MENSURATION  OF  SURFACES. 

(80.)  The  area  of  a  figure  is  the  space  contained  within  the 
line  or  lines  by  which  it  is  bounded.  This  area  is  determined 
by  finding  how  many  times  the  figure  contains  some  other  sur- 
face, which  is  £|,ssumed  as  the  unit  of  measure.  This  unit  is 
commonly  a  square  ;  such  as  a  square  inch,  a  square  foot,  a 
square  rod,  &c.  •  •■ 

The  superficial  unit  has  generally  the  same  name  as  the 
linear  unit,  which  forms  the  side  of  the  square.     Thus, 
the  side  of  a  square  inch  is  a  linear  inch  ; 
"      "    of  a  square  foot  is  a  linear  foot ; 
"      "    of  a  square  yard  is  a  linear  yard,  &o. 

There  are  some  superficial  units  which  have  no  correspond- 
ing linear  units  of  the  same  narhe,  as,  for  example,  an  acre. 

The  following  table  contains  the  square  measures  in  com- 
mon use : 

Table  of  Square  Measures. 

8q.  Inches.  Sq.  Feet. 

144=  1  Sq.  Yards. 

1296=  9    =  1  Sq.Rods. 

39204=  2721=  30|=  1    ^^,,, 

627264=        4356  =        484  =        16=       1     j„„ 
6272640=      43560  =      4840  =       160=     10=     1   ju 
4014489600=27878400  =3097600  =102400=6400=640=1 

•Problem  I. 
(81.)  To  find  the  area  of  a  parallelogram. 

Rule  I. 
Multiply  the  base  by  the  altitude. 

For  the  demonstration  of  this  rule,  see  Greometry,  Prop.  5 
B.  lY. 


60  Trigonometry. 

Ex.  1.  What  is  the  area  of  a  parallelogram  whose  base  is 
17.5  rods,  and  the  altitude  13  rods  ?  ^  ( 1^<' 

Ans.j  227.5  square  rods.  p(^* 

Ex.  2,  What  is  the  area  of  a  square  whose  side  is  315.7 /"'~' 
feet?  ^?^5.,  99666.49  square  feet. 

Ex.  3.  What  is  the  area  of  a  rectangular- board  whose  length 
is  15.25  feet,  and  breadth  15  inches  ? 

Ans.,  19.0625  square  feet. 
Ex.  4.  How  many  square  yards  are  there  in  the  four  sides 
of  a  room  which  is  18  feet  long,  15  feet  broad,  and  9  feet  high  ? 

Ans.f  66  square  yards. 
(82.)  If  the  sides  and  angles  of  a  parallelogram  are  given, 

the  perpendicular  height  may  be  found  by         j^ q 

Trigonometry.  For  DE  is  one  side  of  a 
right-angled  triangle,  of  which  AD  is  the 
hypothenuse.     Hence, 

E  :  AD  :  :  sin.  A  :  DE  ; 

-           ,  .  ,                    ^^    ADXsin.  A 
from  which  DE= r^ . 

mi,      r       ^r               a  r> w TM^    ABxADXsin.  A 
Therefore,  the  area= ABxDE= ^ . 

Hence  we  derive 

Rule  II. 

Multiply  together  two  adjacent  sides,  and  the  sine  of  the 
included  angle. 

Ex.  1.  What  is  the  area  of  a  parallelogram  having  an  angle 
of  58°,  and  the  including  sides  36  and  25.5  feet  ? 

Ans.  The  area=36x 25.5 X. 84805  (natural  sine  of  58°)== 
778.508  square  feet. 

The  computation  will  generally  be  most  conveniently  per- 
formed by  logarithms. 

Ex.  2.  What  is  the  area  of  a  rhombus,  each  of  whose  sides 
is  21  feet  3  inches,  and  each  of  the  acute  angles  53°  20'  ? 

Ans.,  362.209  feet. 

Ex.  3.  How  many  acres  are  contained  in  a  parallelogram 
one  of  whose  angles  is  30°,  and  the  including  sides  are  25.35 
and  10.4  chains  ?  "Ans.,  13  acros,  29.12  rods 


Mensuration  of  Surfaces.        61 

Problem  II. 
(83.)  To  find  the  area  of  a  triangle. 

Rule  I. 

Multiply  the  base  by  half  the  altitude. 

For  demonstration,  see  Greometry,  Prop.  6,  B.  IV. 

Ex.  1.  How  many  square  yards  are  contained  in  a  triangle 
whose  base  is  49  feet,  and  altitude  25 J  feet  ? 

Ans.,  68.736. 

Ex.  2.  What  is  the  area  of  a  triangle  whose  base  is  45  feet, 
and  altitude  27.5  feet  ?  Ans.^  618.75  square  feet. 

(84.)  When  two  sides  and  the  included  angle  are  given,  we 
may  use 

Rule  II. 
Multiply  half  the  product  of  two  sides  by  the  sine  of  the 
included  angle. 

The  reason  of  this  rule  is  obvious,  from  Art.  82,  since  a  tri- 
angle is  half  of  a  parallelogram,  having  the  sstme  base  and  al- 
titude. 

Ex.  1.  What  is  the  area  of  a  triangle  of  which  two  sides  are 
45  and  32  feet,  and  the  included  angle  46°  30'  ? 

Ans.  The  area= 45 X 16 X. 725374  (natural  sine  of  46°  30')=- 

522.269  feet. 
Ex.  2.  What  is  the  area  of  a  triangle  of  which  two  sides  are 
127  and  96  feet,  and  the  included  angle  67°  15'  ? 

Ans. 
(85.)  Wlian  the  three  sides  are  known,  we  may  use 

Rule  III. 
From  half  the  sum  of  the  three  sides  subtract  each  side  seV' 
erally  ;  multiply  together  the  half  sum  and  the  three  remain* 
ders,  and  extract  the  square  root  of  the  product. 

Demonstration. 
Let  a,  by  c  denote  the  sides  of  the  tri- 
angle ABC  ;  then,  by  Greometry,  Prop.  12, 
B.  IV.,  we  have  BC'=AB'+AC'-2ABx 
\J),  or  a''=b'-hc^-2cXA'D  ;  whence,  A. 


62  Trigonometry 

2c 
But  CD'=AC^-AD»; 

hence      Gl>'=l>'J^±^J-^^:ii^t^^, 

or  CD= )z -. 

2c 

ABxCD 


But  the  area= ^ =iVWc'-(b'+c'-ay. 

The  quantity  under  the  radical  sign  being  the  difference  ol 
two  squares,  may  be  resolved  into  the  factors  2bc+(b^+c^—a') 
and  2bc—(b^+c'^—a'');  and  these,  in  the  same  manner,  may 
be  resolved  into        (b+c+a)x(b+c—a)f 
and  (a+b—c)X(a—b+c). 

Hence,  if  we  put  S  equal  to  — ^ — ,  we  shall  have 

the  area=  VS(S-a)  (S-b)  (S-c). 
Ex.  1.  Whatsis  the  area  of  a  triangle  whose  sides  are  125, 
173,  and  216  feet  ? 

Here  S=257,  S-^»=84, 

S-a=132j  S-c=41. 

Hence  the  area=  V257 X 132 X 84 X 41 = 10809  square  feet. 
Ex.  2.  How  many  acres  are  contained  in  a  triangle  whose 
sides  are  49,  50.25,  and  25.69  chains  ? 

Ans.f  61  acres,  1  rood,  39.68  perches. 
Ex.  3.  What  is  the  area  of  a  triangle  whose  sides  are  234, 
289,  and  345  feet? 

Ans. 
(86.)  In  an  equilateral  triangle,  one  of  whose  sides  is  a,  the 
expression  for  the  area  becomes 

_flV3 

~    4     ^ 
that  is,  the  area  of  an  equilateral  triangle  is  equal  to  on 
fourth  the  square  of  one  of  its  sides  multiplied  by  the  square 
root  of  3. 

Ex.  What  is  the  area  of  a  triangle  whose  sides  are  each  37 
feet?  Ans.,  592.79  feet 


( 


Meiisuration   of   Surfaces.  63 

Problem  III. 
(87.)  To  find  the  area  of  a  trapezoid. 

Rule.  ^ 

Multiply  half  the  sum  of  the  parallel  sides  into  their  per 
pendicular  distance. 

For  demonstration,  see  Greometry,  Prop.  7,  B.  IV. 

Ex.  1.  What  is  the  area  of  a  trapezoid  whose  parallel  sidea 
are  156  and  124,  and  the  perpendicular  distance  between  them 
57  feet? 

Ans.,  7980  feet. 

Ex.  2.  How  many  square  yards  in  a  trapezoid  whose  par- 
allel sides  are  678  and  987  feet,  and  altitude  524  feet? 

Ans. 

Problem  IY. 
(88.)  To  find  the  area  of  an  irregular  polygon. 

Rule. 
Draw  diagonals  dividing  the  polygon  into  triangles^  and 
find  the  sum  of  the  areas  of  these  triangles. 

Ex.  1.  What  is  the  area  of  a  quadrilateral,  one  of  whose 
diagonals  is  126  feet,  and  the  two  perpendiculars  let  fall  upon 
i^  from  the  opposite  angles  are  74  and  28  feet  ? 

Ans.,  6426  feet. 
Ex.  2.  In  the  polygon  ABODE,  there  p 

are  given  EC=205,  EB=242,  AF=65, 
CG-=114,  and  DH=110,  to  find  the  area.  X       \^C 

Ans.  ^/^'""^     "^^ 


(89.)  If  the  diagonals  of  a  quadrilateral 
are  given,  the  area  may  be  found  by  the 
following 

Rule. 
Multiply  half  the  product  of  the  diagonals  by  the  sine  of 
the  angle  at  their  intersection. 

Demonstration. 
The  sines  of  the  four  angles  at  E  are  all  equal  tc  each  other 


64 


Trigonometry. 


since  the  adjacent  angles  AED,  DEC  are  the  supplements  of 
each  other  (Art.  27).     But,  according  to        jj 
the  Rule,  Art.  84,  the  area  of 
the  triangle  A|>E=iAExBEXsine  E 

"         "       AED=jAExDEXsineE 

"         "       BEC^JBExECXsineE 

"         ^«       DEC=iDExECXsineE. 
Thereffore, 

the  area  of  ABCD=|(AE+EC)x(BE+ED)Xsine  E 
=JACxBDxsine  E. 

Ex.  1.  If  the  diagonals  of  a  quadrilat^al  are  34  and  56 
rods,  and  if  tlfey  intersect  at  an  angle  of  67°,  what  is  the  area? 

Ans.,  876.32. 

Ex.  2.  If  the  diagonals  of  a  quadrilateral  are  75  and  49, 
and  the  angle  of  intersection  is  42°,  what  is  the  area  ? 


Ans. 


Problem  V. 
(90.  j  To  find  the  area  of  a  regular  polygon. 


Rule  I.  '  • 

Multiplij  half  the  perimeter  by  the  perpendicular  let  fall 
from  the  center  on  one  of  the  sides. 
For  demonstration,  see  Greometry,  Prop.  7,  B.  VI. 
Ex.  1.  What  is  the  area  of  a  regular  pentagon  whose  side 
is  26y  and  the  perpendicular  from  the  center  17.205  feet  ? 

Ans.,  1075.31  feet. 
Ex.  2.  What  is  the  area  of  a  regular  octagon  whose  side  is 
53,  and  the  perpendicular  63.977  ? 

^  Ans. 

^91.)  When  the  perpendicular  is  not  given,  it  may  be  com- 
puted from  the  perimeter  and  number  of  sides.  If  we  divide 
360  degrees  by  the  number  of  sides  of  the 
polygon,' the  quotient  will  be  the  angle  ACB 
at  the  center,  subtended  by  one  of  the  sides. 
The  perpendicular  CD  bisects  the  side  AB, 
and  the  angle  ACB.  Then,  in  the  triangle 
ACD,  we  have  (Art.  42), 

R  :  AD  :  cot.  ACD  :  CD  ;  that  is, 


Mens    jiationsof   Surfaces.  65 

Radius  is  to  half  of  one  of  the  sides  of  the  polygon^  as  the 
cotangent  of  the  opposite  angle  is  to  the  perpendicular  from 
the  center. 

Ex.  3.  Find  the  area  of  a  regular  hexagon  whose  side  is  32 
inches. 

The  angle  ACD  is  j\  of  3160^=30°.     Then 
R  :  16  : :  cot.  30°  :  27.7128= CD,  the  perpendicular ; 
and  the  area=27.7128x  16x6 =2660.4288. 

Ex.  4.  Find  the  area  of  a  regular  decagon  whose  side  is  46 
feet.  Ans.,  16280.946. 

(92.)  In  this  manner  was  computed  the  following  table  of 
the  areas  of  regular  polygons,  in  which  the  side  of  each  poly- 
g«tXL  is  supposed  to  be  a  unit. 

Table  of  Regular  Polygons. 


Names. 

Sides. 

Areas. 

Triangle, 

3 

0.4330127. 

i  Square, 

4 

1.0000000. 

J^  Pentagon, 

5 

1.7204774. 

^  n  Hexagon, 

6 

2.5980762. 

3  Heptagon, 

7 

3.6339124 

^Octagon, 

8 

4.8284271 

^  Nonagon, 

9 

6.1818242. 

f  Decagon, 

10 

7.6942088. 

TUndecagon, 

11 

9.3656399. 

,r^  Dodecagon, 

12 

11.1^61524. 

By  the  aid  of  this  table  may  be 

computed  the  area  of  any 

other  regular  polygon 

having  not  more  than  twelve  sides.    For, 

since  the  areas  of  similar  polygons 

are  as  the  squares  of  their 

homologous  sides,  we 

derive 

Rule  II. 
Multiply  the  square  of  one  of  the  sides  of  the  polygon  by 
the  area  of  a  similar  polygon  whose  side  is  unity. 

Ex.  5.  What  is  the  area  of  a  regular  nonagon  whose  side 
is  63  ?  Ans.,  24535.66. 

Ex.  6.  "What  is  the  area  of  a  regular  dodecagon  whose  side 
is  54  feet?  Ans.,  32647.98  feet. 

E 


6?>  Trigonometry. 

Problem  VI. 
(93.)  To  find  the  circumference  of  a  circle  from  its  diameter 

Rule. 
Multiply  the  diameter  by  3.14159. 

For  the  demonstration  of  this  rule,  see  Greometry,  Prop.  13, 
Cor.  2,  B.VI. 

When  the  diameter  of  the  circle  is  small,  and  no  great  ac- 
curacy is  required,  it  may  be  sufficient  to  employ  the  value 
of  TT  to  only  4  or  5  decimal  places.  But  if  the  diameter  is 
large,  and  accuracy  is  required,  it  will  be  necessary  to  employ 
a  corresponding  number  of  decimal  places  of  tt.  The  value  of 
TT  to  ten  decimal  places  is  3.14159,26536, 

and  its  logarithm  is  0.497150. 
Ex.  1.  What  is  the  circumference  of  a  circle  whose  diame- 
ter is  125  feet? 

Ans.,  392.7  feet. 
Ex.  2.  If  the  diameter  of  the  earth  is  7912  miles,  what  is 
its  circumference  ? 

Ans.,  24856.28  miles. 
Ex.  3.  If  the  diameter  of  the  earth's  orbit  is  189,761,000 
miles,  what  is  its  circumference  ? 

Ans.,  596,151,764  miles. 
To  obtain  this  answer,  the  value  of  ir  must  be  taken  to  at 
least  eight  decimal  places. 

^  Problem  VII. 

(94.)  To  find  the  diameter  of  a  circle  from  its  circum^ 
ference. 

Rule  I. 

Divide  the  circumference  by  3.14159. 

This  rule  is  an  obvious  consequence  from  the  preceding 
To  divide  by  a  number  is  Jthe  same  as  to  multiply  by  its  re- 
ciprocal;  and,  since  multiplication  is  more  easily  performed 
than  division,  it  is  generally  most  convenient  to  multiply  by 
the  reciprocal  of  tt,  which  is  0.3183099.     Hence  we  have 

Rule  II. 
Multiply  the  circumference  by  0.31831. 


Mensuration  of  Surfaces.        67 

Ex.  1.  What  is  the  diameter  of  a  circle  whose  circumference 
IS  875  feet  ? 

Ans.,  278.52  feet. 
Ex.  2.  If  the  circumference  of  the  moon  is  6786  miles,  what 
is  its  diameter  ? 

Ans.,  2160  miles. 
Ex.  3.  If  the  circumference  of  the  moon's  orbit  is  1,492,987 
miles,  what  is  its  diameter  ? 

Ans.,  475,233  miles. 

Problem  YIII. 
(95.)  To  find  the  length  of  an  arc  of  a  circle. 

BULE   I. 

As  360  is  to  the  number  of  degrees  in  the  arc^  so  is  the  cir- 
cumference  of  the  circle  to  the  length  of  the  arc. 

This  rule  follows  from  Prop.  14,  B.  III.,  in  Greometry,  where 
it  is  proved  that  angles  at  the  center  of  a  circle  have  the  same 
ratio  with  the  intercepted  arcs. 

Ex.  1.  AVhat  is  the  length  of  an  arc  of  22°,  in  a  circle  whose 
diameter  is  125  feet  ? 

The  circumference  of  the  circle  is  found  to  be  392.7  feet. 
Then  360  :  22  : :  392.7  :  23.998  feet. 

Ex.  2.  If  the  circumference  of  the  earth  is  24,856.28  miles 
what  is  the  length  of  one  degree  ? 

Ans.,  69.045  miles. 

Rule  II. 

(96.)  Multiply  the  diameter  of  the  circle  by  the  number  of 
degrees  in  the  arc,  and  this  product  by  0.0087266. 

Since  the  circumference  of  a  circle  whose  diameter  is  unity 
is  3.14159,  if  we  divide  this  number  by  360,  we  shall  obtain 
the  length  of  an  arc  of  one  degree,  viz.,  0.0087266.  If  we 
multiply  this  decimal  by  the  number  of  degrees  in  any  arc,  we 
shall  obtain  the  length  of  that  arc  in  a  circle  whose  diameter 
is  unity ;  and  this  product,  multiplied  by  the  diameter  of  any 
other  circle,  will  give  the  length  of  an  arc  of  the  given  num- 
ber of  degrees  in  that  circle. 


68  Trigonometry. 

Ex.  3.  What  is  the  length  of  an  arc  of  25°,  in  a  circle  whose 
radius  is  44  rods  ? 

Ans.,  19.198  rods. 
Ex.  4.  "What  is  the  length  of  an  arc  of  11°  15',  in  a  circle 
whose  diameter  is  1234  feet  ? 

Ans.,  121.147  feet. 
(97.)  If  the  number  of  degrees  in  an  arc  is  not  given,  it  may 
be  comjmted  from  the  radius  of  the  circle,  jj 

and  either,  the  chord  or  height  of  the  arc. 
Thus,  let  AB  be  the  chord,  and  DE  the   ^/ 
height  of  the  arc  ADB,  and  C  the  center 
of  the  circle.    Then,  in  the  right-angled  tri- 
angle ACE, 

.p  .P  .     (  AE  :  sin.  ACE, 
^^•^••(CE:  cos.  ACE, 
either  of  which  proportions  will  give  the  number  of  degrees  in 
half  the  arc. 

If  only  the  chord  and  height  of  the  arc  are  given,  the  diam- 
eter of  the  circle  may  be  found.  For,  by  Greometry,  Prop.  22, 
Cor.,  B.  IV., 

DE  :  AE  : :  AE  :  EF. 
Ex.  5.  What  is  the  length  of  an  arc  whose  chord  is  6  feet, 
in  a  circle  whoso  radius  is  9  feet  ? 

-  Ans.,  -6.117  feet. 

Problem  IX. 
(98.)  To  find  the  area  of  a  circle. 

Rule  I. 
Multiply  the  circumference  by  half  the  radius. 
For  demonstratian,  see  Greometry,  Prop.  12,  B,  VI. 

Rule  II. 
Multiply  the  square  of  the  radius  by  3.14159. 
See  Geometry,  Prop  13,  Cor.  3,  B.  VI. 
Ex.  1.  What  is  the  area  of  a  circle  whose  diameter  is  18 
feet  ? 

Ans.,  254.469  feet. 


Mensuration   of   Surfaces  69 

Ex.  2.  What  is  the  area  of  a  circle  whose  circumference  is 
74  feet  ? 

Ans.,  435.766  feet. 
Ex.  3.  What  is  the  area  of  a  circle  whose  radius  is  125 
yards  ? 

Ans.,  49087.38  yards. 

Problem  X. 
(99.)  To  find  the  area  of  a  sector  of  a  circle. 

Rule  I 
Multiply  the  arc  of  the  sector  by  half  its  radius. 
See  Geometry,  Prop.  12,  Cor.,  B.  VI. 

Rule  II. 
As  360  is  to  the  number  of  degrees  in  the  arc^  so  is  the  area 
of  the  circle  to  the  area  of  the  sector. 

This  follows  from  Greometry,  Prop.  14,  Cor.  2,  B.  III. 
Ex.  1.  What  is  the  area  of  a  sector  whose  arc  is  22°,  in  a 
circle  whose  diameter  is  125  feet  ? 

The  length  of  the  arc  is  found  to  he  23.998. 
Hence  the  area  of  the  sector  is  749.937. 
Ex.  2.  What  is  the  area  of  a  sector  whose  arc  is  25°,  in  a 
circle  whose  radius  is  44  rods  ? 

Ans.,  422.367  rods. 
Ex.  3.  What  is  the  area  of  a  sector  less  than  a  semicircle, 
whose  chord  is  6  feet,  in  a  circle  whose  radius  is  9  feet  ? 

Ans.,  27.522  feet. 

Problem  XI. 
(100.)  To  find  the  area  of  a  segment  of  a  circle. 

Rule. 

Find  the  c^ea  of  the  sector  which  has  the  same  arc,  and 
also  the  area  of  the  triangle  formed  by  the  chord  of  the  seg' 
ment  and  the  radii  of  the  sector. 

Then  take  the  sum  of  these  areas  if  the  segment  is  greater 
than  a  semicircle,  but  take  their  difference  if  it  is  less 


^4^ 


./yO 


70  Trigonometry. 

It  is  obvious  that  the  segment  AEB  is 
equal  to  the  sum  of  the  sector  ACBE  and 
the  triangle  ACB,  and  that  the  segment 
ADB  is  equal  to  the  difference  between 
the  sector  ACBD  and  the  triangle  ACB. 

Ex.  1.  What  is  the  area  of  a  segment 
whose  arc  contains  280°,  in  a  circle  whose 
diameter  is  50  ? 

The  whole  circle      =       1963.495 
,      The  sector  =       1527.163 

The  triangle  =         307.752 

The  segment  =       1834.915,  Ans. 

Ex,  2.  What  is  the  area  of  a  segment  whose  chord  is  20  feet, 
and  height  2  feet  ? 

Ans.,  26.8788  feet. 
Ex.  3.  What  is  the  area  of  a  segment  whose  arc  is  25°,  in 
a  circle  whose  radius  is  44  rods  ? 

Ans. 
(101.)  The  area  of  the  zone  ABHGr,  included  between  two 
parallel  chords,  is  equal  to  the  difference  between  the  segments 
ami  and  ADB. 

Ex.  4.  What  is  the  area  of  a  zone,  one  side  of  which  is  96, 
and  the  other  side  60,  and  the  distance  between  them  26  ? 

Ans.,  2136.7527. 
The  radius  of  the  circle  in  this  example  will  be  found  to 
be  50. 

Problem  XII. 
(102.)  To  find  the  area  of  a  ring  included  between  the  cir 
cumferences  of  two  concentric  circles. 

Rule. 
Take  the  difference  between  the  areas  of  the  two  circles;  or, 
Subtract  the  square  of  the  less  radius  from  the  square  of  the 
greater,  and  multiply  their  difference  by  3.14159. 
For,  according  to  G-eometry,  Prop.  13,  Cor.  3,  B.  VI., 
the  area  of  the  greater  circle  is  equal  to  ir  R', 
and  the  area  of  the  smaller,  ir  r'. 

Their  difference,  or  the  area  of  the  ring,  is  ir  (R^— r'). 


Mensuration   of   Solids.  71 

Ex.  1.  The  diameters  of  two  concentric  circles  are  60  and 
50.  What  is  the  area  of  the  ring  included  between  their  cir- 
cumferences ? 

Ans.,  863.938. 
Ex.  2.  The  diameters  of  two  concentric  circles  are  320  and 
280     What  is  the  area  of  the  ring  included  between  their  cir- 
cumferences ?  ,r   /  £  *2/  (^'  ^ 
\i   \                                    Ans.,  18849.55 

Problem  XIII.  ^ 

(103.)  To  find  the  area  of  an  ellipse,  ,  o  "^ 

Rule. 
Multiply  the  product  of  the  semi-axes  by  3.14159. 
For  demonstration,  see  Greometry,  Ellipse,  Prop.  21.  -^    '  ^ 

Ex.  1.  What  is  the  area  of  an  ellipse  whose  major  axis  ia 
70  feet,  and  minor  axis  60  feet  ? 

Ans.,  3298.67  feet.  va, 

Ex.  2.  What  is  the  area  of  an  ellipse  whose  axes  are  340  |\^"V^ 
and  310?  %/^/     ^'    ^        .  ^ 

;.ty  Ans.,  82780.896 

3  Problem  XIV.  !^  r 

(104.)  To  find  the  area  of  a  parabola. 

^n  0 
Rule.  "^^ 

Multiply  the  base  by  two  thirds  of  the  height. 
For  demonstration,  see  Greometry,  Parabola,  Prop.  12. 
Ex.  1.  What  is  the  area  of  a  parabola  whose  base  is  18  feet, 
and  height  o  feet  ? 

Ans.^  60  feet. 
Ex.  2.  What  is  the  area  of  a  parabola  whose  base  is  525 
feet,  and  height  350  feet  ? 

Ans.,  122500  feet 


MENSURATION  OF  SOLIDS. 
(105.)  The  common  measuring  unit  of  solids  is  a  cube^ 
whose  faces  are  squares  of  the  same  name ;  as,  a  cubic  inch, 
a  cubic  foot,  &c.     This  measuring  unit  is  not,  however,  of 


^■- 


72  .  Trigonometry. 

necessity  a  cube  whose  faces  are  squares  of  the  same  namtj. 
Thus  a  bushel  may  have  the  form  of  a  cube,  but  its  faces  can 
only  be  expressed  by  means  of  some  unit  of  a  different  denom- 
ination.    The  following  is 

The  Table  of  Solid  Measure. 


1728 

cubic  inches 

= 

1  cubic  foot. 

27 

cubic  feet 

= 

1  cubic  yard. 

4492i 

cubic  feet 

= 

1  cubic  rod. 

231 

cubic  inches 

= 

1  gallon  (liquid  measure), 

268.8  cubic  inches 

= 

1  gallon  (dry  measure). 

2150.4  cubic  inches 

= 

1  bushel. 

Problem  I. 
(106.)  To  find  the  surface  of  a  right  prism. 

Rule. 
Multiply  the  perimeter  of  the  base  by  the  altitude  for  the 
convex  surface.     To  this  add  the  areas  of  the  two  ends  when 
the  entire  surface  is  required. 
See  Geometry,  Prop.  1,  B.  VIII. 

Ex.  1.  What,  is  the  entire  surface  of  a  parallelepiped  whose 
altitude  is  20  feet,  breadth  4  feet,  and  depth  2  feet  ? 

Ans.^  256  square  feet. 
Ex.  2.  What  is  the  entire  surface  of  a  pentagonal  prism 
whose  altitude  is  25  feet  6  inches,  and  each  side  of  its  base  3 
feet  9  inches  ? 

Ans.,  526.513  square  feet. 
Ex.  3.  What  is  the  entire  surface  of  an  octagonal  prism 
whose  altitude  is  12  feet  9  inches,  and  each  side  of  its  base  2 
feet  5  inches  ? 

Ans.,  302.898  square  feet. 

Problem  II. 
(107.)  To  find  the  solidity  of  a  prism. 

Rule. 
Multiply  the  area  of  the  base  by  the  altitude. 
See  Geometry,  Prop.  11,  B.  YIII. 


Mensuration   of   Solids  73 

Ex.  1.  What  is  the  solidity  of  a  parallelopiped  whose  alti- 
tude is  30  feet,  breadth  6  feet,  and  depth  4  feet  ? 

Ans..f  720  cubic  feet. 
Ex.  2.  "What  is  the  solidity  of  a  square  prism  whose  altitude 
is  8  feet  10  inches,  and  each  side  of  its  base  2  feet  3  inches  ? 

Ans.,  44||  cubic  feet. 
Ex.  3   "What  is  the  solidity  of  a  pentagonal  prism  whoso  a.- 
titude  is  20  feet  6  inches,  and  its  side  2  feet  7  inches  ? 

Ans,,  235.376  cubic  feet. 

Problem  III. 
(108.)  To  find  the  surface  of  a  regular  pyramid. 

Rule. 
Multiply  the  perimeter  of  the  base  by  half  the  slant  height 
for  the  convex  surface.     To  this  add  the  area  of  the  base 
when  the  entire  surface  is  required. 
See  G-eometry,  Prop.  14,  B.  VIII. 

Ex.  1.  What  is  the  entire  surface  of  a  triangular  pyramid 
whose  slant  height  is  25  feet,  and  each  side  of  its  base  5  feet  ? 

Ans.^  198.325  square  feet. 
Ex.  2.  What  is  the  entire  surface  of  a  square  pyramid 
whose  slant  height  is  30  feet,  and  each  side  of  the  base  4  feet  ? 

Ans.,  256  square  feet. 
Ex.  3.  What  is  the  entire  surface  of  a  pentagonal  pyra- 
mid whose  slant  height  is  20  feet,  and  each  side  of  the  base 
3  feet? 

Ans.^  165.484  square  feet. 

Problem  IY. 
(109.)  To  find  the  solidity  of  a  pyramid. 

Rule. 
Multiply  the  area  of  the  base  by  one  third  of  the  altitude. 
See  Geometry,  Prop.  17,  B.  VIII. 

Ex.  1.  What  is  the  solidity  of  a  triangular  pyramid  whose 
altitude  is  25  feet,  and  each  sid  3  of  its  base  6  feet  ? 

Ans.,  129.904  cubic  feet 


74  Trigonometry. 

Ex.  2.  What  is  the  solidity  of  a  square  pyramid  whose  slam 
height  is  22  feet,  and  each  side  of  its  base  10  feet  ? 

A71S.J  714.143  cubic  feet. 
Ex.  3.  What  is  the  solidity  of  a  pentagonal  pyramid  whaso 
altitude  is  20  feet,  and  each  side  of  its  base  3  feet  ? 

Afis.,  103.228  cubic  feet. 

Problem  Y. 
(110.)  To  find  the  surface  of  a  frustum  of  a  regular  pyr- 
amid. 

Rule.  • 

Multiply  half  the  slant  height  by  the  sum  of  the  perime- 
ters  of  the  two  bases  for  the  convex  surface.  To  this  add 
the  areas  of  the  two  bases  when  the  entire  surface  is  re- 
quired. 

See  G-eometry,  Prop.  14,  Cor.  1,  B.  YIII. 

Ex.  1.  What  is  the  entire  surface  of  a  frustum  of  a  square 
pyramid  whose  slant  height  is  15  feet,  each  side  of  the  greater 
base  being  4  feet  6  inches,  and  each  side  of  the  less  base  2  feet 
10  inches  ? 

Ans.,  248.278  square  feet. 

Ex.  2.  What  is  the  entire  surface  of  a  frustum  of  an  oc- 
tagonal  pyramid  whose  slant  height  is  14  feet,  and  the  sides 
of  the  ends  3  feet  9  inches,  and  2  feet  3  inches  ? 

Ans.^  428.344  square  feet. 

Problem  VI. 
(111.)  To  find  the  solidity  of  a  frustum  of  a  pyramid. 

Rule. 

Add  together  the  areas  of  the  two  bases,  and  a  mean  pro- 
portional  between  them,  and  multiply  the  sum  by  one  third 
of  the  altitude. 

See  G-eometry,  Prop.  18,  B.  VIII. 

Wlien  the  pyramid  is  regular,  it  is  generally  most  conven- 
ient to  find  the  area  of  its  base  by  Rule  II.,  Art.  92.  If  we 
put  a  to  represent  one  side  of  the  lower  base,  and  b  one  side 
of  the  upper  base,  and  the  tabular  number  from  Art.  92  bj 


Mensuration   op    Solids. 


75 


T,  the  area  of  the  lower  base  will  be  a'T  ]  thai  of  the  upper 
base  will  be  ^^T  ;  and  the  mean  proportional  will  be  a^T. 
Hence,  if  we  represent  the  height  of  the  frustum  by  A,  its  so- 
lidity will  be 

Ex.  1.  What  is  the  solidity  of  a  frustum  of  an  hexagonal 
pyramid  whose  altitude  is  15  feet,  each  side  of  the  greater  end 
being  3  feet,  and  that  of  the  less  end  2  feet  ? 

Ans.,  246.817  cubic  feet. 
Ex.  2.  What  is  the  solidity  of  a  frustum  of  an  octagonal 
pyramid  whose  altitude  is  9  feet,  each  side  of  the  greater  end 
being  30  inches,  and  that  of  the  less  end  20  inches  ? 

Ans.,  191.125  cubic  feet. . 


(112) 


Definition. 
A  wedge  is  a  solid  bounded  by  five  planes,  viz.,  a  rec- 


tangular base,  ABCD,  two  trape- 
zoids, ABFE,  DCFB,  meeting  in 
an  edge,  and  two  triangular  ends, 
ADE,  BCF.  The  altitude  of  the 
wedge  is  the  perpendicular  drawn 
from  any  point  in  the  edge  to  the 
plane  of  the  base,  as  EH. 


Problem  YH. 
(113.)  To  find  the  solidity  of  a  wedge. 

Rule. 
Add  the  length  of  the  edge  to  twice  the  length  of  the  base^ 
and  multiply  the  sum  by  one  sixth  of  the  product  of  the  height 
of  the  wedge  and  the  breadth  of  the  base. 

Demonstration. 
Put  L=AB,  the  length  of  the  base ; 
"     /=EF,  the  length  of  the  edge  ; 
"     6=BC,  the  breadth  of  the  base ; 
"     ^=EH,  the  altitude  of  the  wedge. 
Now,  if  the  length  of  the  base  is  equal  to  that  of  the  edge, 


f6  Trigonometry. 

it  is  evident  that  the  wedge  is  half  of  a  pri&m  of  the  same  base 
and  height.  If  the  length  of  the  base  is  greater  than  that  of 
the  edge,  let  a  plane,  EQI,  be  drawn  parallel  to  BCF.  The 
wedge  will  be  divided  into  two  parts,  viz.,  the  pyramid  E  — 
AIGrD,  and  the  triangular  prism  BCF— Gr. 

The  solidity  of  the  former  is  equal  to  iM(L— Z),  and  that 
of  the  latter  is  \bhl.     Their  sum  is 

If  the  length  of  the  base  is  less  than  that  of  the  edge,  the 
wedge  will  be  equal  to  the  difference  between  the  prism  and 
pyramid,  and  we  shall  have 

iMZ-iM(Z-L), 
which  is  equal  to 

\bhl+\bh{l.'-l), 
the  same  result  as  before. 

Ex.  1.  "What  is  the  solidity  of  a  wedge  whose  base  is  30 
inches  long  and  5  inches  broad,  its  altitude  12  inclies,  and  the 
length  of  the  edge  2  feet  ? 

Ans.,  840  cubic  inches. 
Ex.  2.  "What  is  the  solidity  of  a  wedge  whose  base  is  40 
inches  long  and  7  inches  broad,  its  altitude  18  inches,  and  the 
length  of  the  edge  30  inches  ? 

Ans.<)  2310  cubic  inches. 

Definition, 
(114.)  A  rectangular  prismoid  is  a  solid  bounded  by  six 
planes,  of  which  the  two  bases  are  rectangles  having  their  cor- 
responding sides  parallel,  and  the  four  upright  sides  of  the  sol- 
id are  trapezoids. 

Problem  YIII. 
To  find  the  solidity  of  a  rectangular  prismoid. 

Rule. 
Add  together  the  areas  of  the  two  bases ^  and  four  times  the 
area  of  a  parallel  section  equally  distant  from  the  bases,  and 
multiply  the  sum  by  one  sixth  of  the  altitude. 

Demonstration. 
Put  L  and  B= length  and  breadth  of  one  base ; 


Mensuration   of    Solids.  77 

Put  /  and  b  =  length  and  breadth  of  the  other  base; 

*•   M    "   m= length  and  breadth  of  middle  sec;    r—r {. 

"    h  =the  altitude  of  the  prismoid.  /  \ — I — \-\ 

It  is  evident  that  if  a  plane  be  made  to  pass  km  \  \    \ 

through  the  opposite  edges  of  the  upper  and  I   N -^  \    \ 

lower  bases,  the  prismoid  will  be  divided  into  \B  1  \  \ 

two  wedges,  whose  bases  are  the  bases  of  the     ^^ — ^ 

prismoid,  and  whose  edges  are  L  and  /.     The  solidity  of  these 
wedges,  and,  consequently,  that  of  the  prismoid,  is 

iB/i(2L+/)+iM(2Z+L)=p(2BL+BZ+26^+6L). 
But,  since  M  is  equally  distant  from  L  and  /,  we  have 

2M=L+/,  and  2m=B+6;  ^ 

hence         4Mm=(L+/)  (B+^>)=BL+B/+^>L+R 

Substituting  4Mm  for  its  value  in  the  preceding  expression, 
we  obtain  for  the  solidity  of  the  prismoid 
iA(BL+Z^Z+4Mm). 
Ex.  1.  What  are  the  contents  of  a  log  of  wood,  in  the  form 
of  a  rectangular  prismoid,  the  length  and  breadth  of  one  end 
oeing  16  inches  and  12  inches,  and  of  the  other  7  inches  and 
4  inches,  the  length  of  the  log  being  24  feet  ? 

Ans..  16J  cubic  feet. 
Ex.  2.  What  is  the  solidity  of  a  log  of  hewn  timber,  whose 
mds  are  18  inches  by  15,  and  14  inches  by  llj,  its  length  be- 
ing 18  feet?  Ans.,  26fi  cubic  feet. 

Problem  IX. 

To  compute  the  excavation  or  embankment  for  a  rail-way. 

(115.)  By  the  preceding  rule  may  be  computed  the  amount 
of  excavation  or  embankment  required  in  constructing  a  rail- 
load  or  canal.  If  we  divide  the  line  of  the  road  into  portions 
10  small  that  each  may  be  regarded  as  a  straight  line,  and 
suppose  an  equal  number  of  transverse  sections  to  be  made, 
the  excavation  or  embankment  between  two  sections  may  be 
regarded  as  a  prismoid,  and  its  contents  found  by  the  pre 
ceding  rule. 

Let  ABCD  represent  the  lower  surface  of  the  supposed  ex- 
cavation, which  we  will  assume  to  be  parallel  to  the  horizon ; 
and  let  EFG-H  represent  the  upper  surface  of  the  excavation 


78 


Trigonometry. 


projected  ou  a  horizontal  plane.     Also,  let  E'A'B'F',  G'C'D'H 
represent  the  vertical  sections  at         g^    l  m    h' 

the  extremities.  If  we  suppose  ver- 
tical planes  to  pass  through  the  lines 
AC,  BD,  the  middle  part  of  the  ex- 
cavation, or  that  contained  between 
these  vertical  planes,  will  he  a  rect- 
angular prismoid,  of  which  A'B'KI 
will  be  one  base,  and  C'D'ML  the 
other  base.  Its  solidity  will  there- 
fore be  given  by  Art.  114.  The 
parts  upon  each  side  of  the  middle  prismoid  are  also  halves  of 
rectangular  prismoids ;  or,  if  the  two  parts  are  equal,  they 
may  be  regarded  as  constituting  a  second  prismoid,  one  of 
whose  bases  is  the  sum  of  the  triangles  A'E'I,  B'F'K  ;  and  the 
other  base  is  the  sum  of  the  triangles  C'Gr'L,  D'H'M.  There- 
fore the  volume  of  the  entire  solid  is  equal  to  the  product  of 
one  sixth  of  its  length,  by  the  sum  of  the  areas  of  the  sections 
at  the  two  extremities,  and  four  times  the  area  of  a  parallel 
and  equidistant  section. 

Ex.  1.  Let  ABCDEFGr  represent  the  profile  of  a  tract  of 


land  selected  for  the  line  of  a  rail- way ;  and  suppose  it  is  re- 
quired, by  cutting  and  embankment,  to  reduce  it  from  its  pres- 
ent hilly  surface  to  one  uniform  slope  from  the  point  A  to 
the  point  Gr. 

The  distance  AH  is  561  feet ;  the  distance  DK  is  820  feet; 

"  "       HI  is  858  feet;     "         "       KL  is  825  feet; 

"'         "       ID    is  825  feet;     "         "       LG  is  3'JO  feet. 
The  perpendicular  BH  is  18  feet;  the  perpendicular  KE  is  19  feet; 


"  "  CI  is 20 feet; 

The  annexed  figure  repre- 
sents a  cross  section,  showing 
the  form  of  the  excavation. 

The  base  of  the  cutting  is  to 


LFif^  8  feet 


~ 


Mensuration  op    Sclids.  79 

be  50  feet  wide,  the  slope  IJ  horizontal  to  1  perpendicular; 
that  is,  where  the  depth  ad  is  10  feet,  the  width  of  the  slope 
cd  at  the  surface  will  be  15  feet. 

Calculation  of  the  portion  ABH. 
Since  BH  is  18  feet,  the  length  of  cd  in  the  cross  section 
will  be  27  feet,  and  c/,  the  breadth  at  the  top  of  the  section, 
will  be  104  feet.     "We  accordingly  find,  by  Art.  87,  the  area  of 
the  trapezoid  forming  the  cross  section  at  BH  equal  to 

i^Xl8=1386fcet. 

For  the  middle  section,  the  height  is  9  feet,  cd  is  13.5  feet, 
and  cf  is  77  feet.  The  area  of  the  cross  section  is  therefore 
equal  to 

— ^X  9=571.5. 

The  solid  ABH  will  therefore  be  equal  to 

561 
(1386+4x571.5)  -^-=343332  cubic  feet,  or 

12716  cubic  yards. 

Calculation  of  the  portion  BCIH. 
Since  CI  is  20  feet,  the  length  of  cd  is  30  feet,  and  cf  is  1  lO 
feet.     The  area  of  the  section  at  CI  is  therefore  equal  to 

H^X20=1600. 

For  the  middle  section,  the  height  is  19  feet,  cd  is  28.5  fuet, 
and  cf  is  107  feet.  The  area  of  the  cross  section  is  therefore 
equal  to 

107+50  ,^  ,,^,       ;^ 

— X 19=  1491.5. 

The  solid  BCIH  will  therefore  be  equal  to 
(1386+1600+4x1491.5)  -^-=1280136  cubic  feet,  or 
47412.4  cubic  yards. 

Calculation  of  the  portion  CID. 
The  height  of  the  middle  section  is  10  feet ;  therefore  cf  if 
80  feet,  and  the  area  of  the  cross  section  is 


80  Trigonometry. 

80+50 
—^X  10=650. 

The  solid  CID  will  therefore  be  equal  to, 

825 
(1600+4x650)  -g-=577500  cubic  feet,  oi 

21388.9  cubic  yards. 
The  entire  amount  of  excavation  therefore  is, 
ABH=  12716.0  cubic  yards. 
BCIH=47412.4 
CDI=21388.9  " 

Total  excavation,    8151^.3  " 


The  following  is  a  cross  section,  showing  the  form  of  the  em- 
bankment. 

The  top  of  the  embankment  a "b 

is  to  be  50  feet  wide,  the  slope 
2  to  1 ;  that  is,  where  the  height 
ad  is  10  feet,  the  base  cd  is  to  ^ 
be  20  feet. 

Calculation  of  the  portion  DKE. 
Since  EK  is  19  feet,  the  length  of  cd  is  38  feet,  and  cf  is 
126  feet.     The  area  of  the  cross  section  at  EK  is  therefore 
equal  to 

^X19=1672. 

For  the  middle  section,  the  height  is  9.5  feet ;  cd  is  therefore 
19  feet,  and  cf  is  88  feet.  The  area  of  the  cross  section  is 
therefore 

??±^X9.5=655.5 

The  solid  DKE  is  therefore  equal  to 

820 
(1672+4X 655.5) -g-=586846.7  cubi<3  feet,  or  . 

21735.1  cubic  yards. 

Calculation  of  the  portion  KBFL. 
Since  LF  is  8  feet,  cd  is  16  feet,  and  cf  is  82  feet.     The 
area  of  the  section  at  LF  is  therefore  equal  to 


Mensuration  v.f   Solids.  HI 

82+50 
2fZ^X  8=528. 

The  height  of  the  middle  section  is  13.5  feet  •  therefore  cd 
is  27  feet,  and  cf  is  104  feet.  The  area  of  the  crosi^'  section  is 
therefore  equal  to 

104+50 
Y     X  13.5=1039.5. 

The  solid  KEFL  will  therefore  be  equal  to 

(1672+528+4xl039.5)-g-=874225  cubic  feet,  or 

32378.7  cubic  yards. 

Calculation  of  the  portion  LFG-. 

The  height  of  the  middle  section  is  4  feet-;  therefore  cf  is  66 

'eet,  and  the  area  of  the  cross  section  is  equal  to 

66+50     ,     ^^^ 
^      X  4=232. 

The  solid  LFO  will  therefore  be  equal  to 

330 
(528+4x232)-g-=80080  cubic  feet,  or     , 

2965.9  cubic  yards. 
The  entire  amount  of  embankment  therefore  is 
DKE =21735.1  cubic  yards. 
KEFL=32378.7 
LFa=  2965.9 

Total  embankment,  57079.7  '' 

Ex.  2.  Compute  the  amount  of  excavation  of  the  hill  A.BCD 
from  the  following  data : 

The  distance  AH  is  325  feet ;  the  perpendicular  BH  is  12  feet ; 
"  ''       HI  is  672  feet;     "  "  CI  is  13  feet. 

"  "       ID  is  534  feet. 

The  base  of  the  cutting  to  be  50  feet  wide,  and  the  slope  IJ 
aorizontal  to  1  perpendicular.  Ans.,  33969  cubic 'yards. 

Problem  X. 
(116.)  To  find  the  surface  of  a  re^ula,r  polyedron. 

Rule. 
Multiply  the  area  of  one  of  the  faces  by  the  numhei  oj 


82  Trigonometry. 

faces;  or,  Multiptg^ Hie  square  of  one  of  the  edges  by  the 
surface  of  a  similar  solid  whose  edge  is  unity. 

Since  all  the  faces  of  a  regular  polyedron  are  equal,  it  is 
evident  that  the  area  of  one  of  them,  multiplied  by  their  num- 
ber, will  give  the  entire  surface.  Also,  regular  solids  of  the 
same  name  are  similar,  and  similar  polygons  are  as  the  squares 
of  their  homologous  sides  (Geom.^  Prop.  26,  B.  lY.).  The  fol- 
ilowing  table 'shows  the  surface  and  solidity  of  regular  poly- 
hedrons whose  edge  is  unity.  The  surface  is  obtained  by  mul- 
tiplying the  area  of  one  of  the  faces,  as  given  in  Art.  92,  by 
the  number  of  faces.  Thus  the  area  of  an  equilateral  trian- 
gle, whose  side  is  1,  is  0.4830127.  Hence  the  surface  of  a 
regular  tetraedron 

=  .4330127  X  4= 1.7320508, 
and  so  on  for  the  other  solids. 

A  Table  of  the  regular  Polyedrons  whose  Edges  are  unity 

Names.  No.  of  Faces.  Surface.  Solidity. 

Tetraedron,  4  1.7320508         0.1178513. 

Hexaedron,  6  6.0000000         1.0000000. 

Octaedron,  8  3.4641016         0.4714045. 

Dodecaedron,        12         20.6457288         7.6631189. 
Icosaedron,  20  8.6602540         2.1816950. 

Ex.  1.  What  is  the  surface  of  a  regular  octaedron  whose 
edges  are  each  8  feet  ? 

Ans.,  221.7025  feet. 
Ex.  2.  What  is  the  surface  of  a  regular  dodecaedron  whose 
edge  is  12  feet? 

Ans.,  2972.985  feet. 

Problem  XI. 
(117.)  To  find  the  solidity  of  a  regular  polyedron. 

Rule. 

Multiply  the  surface  by  one  third  of  the  perpendicular  let 
fall  from  the  center  on  one  of  the  faces :  or,  Multiply  the 
cube  of  one  of  the  edges  by  the  solidity  of  a  similar  polyedron^ 
whose  edge  is  unity. 

Since  the  faces  of  a  rs^^ular  polyedron  are  similar  and  equal, 


Mensuration  of   Solids.  ^3 

and  the  solid  angles  are  all  equal  to  each  other,  it  is  evident 
that  the  faces  are  all  equally  distant  from  a  point  in  the  solid 
called  the  center.  If  planes  be  made  to  pass  through  the  cen- 
ter and  the  several  edges  of  the  solid,  they  will  divide  it  into 
as  many  equal  pyramids  as  it  has  faces.  The  base  of  each 
pyramid  will  be  one  of  the  faces  of  the  polyedron ;  and  since 
their  altitude  is  the  perpendicular  from  the  center  upon  one  of 
the  faces,  the  solidity  of  the  polyedron  must  be  equal  to  the 
areas  of  all  the  faces,  multiplied  by  one  third  of  this  perpen- 
dicular. 

Also,  similar  pyramids  are  to  each  other  as  the  cubes  ot 
their  homologous  edges  (Geom.,  Prop.  17,  Cor.  3,  B.  VIII.). 
And  since  two  regular  polyedrons  of  the  same  name  may  be 
divided  into  the  same' number  of  similar  pyramids,  they  must 
be  to  each  other  as  the  cubes  of  their  edges. 

(118.)  The  solidity  of  a  tetraedron  whose  edge  is  unity,  may 
be  computed  in  the  following  manner  : 

Let  C— ABD  be  a  tetraedron.  From  one  angle,  C,  let  fall 
a  perpendicular,  CE,  on  the  opposite  face;  c 

draw  EF  perpendicular  to  AD;  and  join 
CF,  AE.  Then  AEF  is  a  right-angled  tri- 
angle, in  which  EF,  being  the  sine  of  30°, 
is  one  half  of  AE  or  BE  ;  and  therefore 
FE  is  one  third  of  BF  or  CF.  Hence  the 
oosine  of  the  angle  CFE  is  equal  to  ^ ;  that   ^v  B 

is,  the  angle  of  inclination  of  the  faces  of  the  polyedron  is  70^ 
31'  U".  Also,  in  the  triangle  CAF,  CF  is  the"  sine  of  60°, 
which  is  0.866025.  Hence,  in  the  right-angled  triangle  CEF, 
knowing  one  side  and  the  angles,  we  can  compute  CE,  which 
is  found  to  be  0.8164966.  Whence,  knowing  the  base  ABD 
(Art.  92),  we  obtain  the  solidity  of  the  tetraedron  =0.1178513. 

In  a  somewhat  similar  manner  may  the  solidities  of  the 
other  regular  polyedrons,  given  in  Art.  116,  be  obtainec? 

Ex.  1.  What  is  the  solidity  of  a  regular  tetraedron  whose 
edges  are  each  24  inches  ? 

Ans.,  0.9428  feet. 

Ex.  2.  What  is  the  scdidity  of  a  regular  icosaedron  whoso 
3dges  are  each  20  feet  ? 

Ans.,  17453.56  feet 


^4  Trigonometry. 

THE  THREE  ROUND  BODIES. 

Problem  I. 
(119.)  To  find  the  surface  of  a  cylinder. 

Rule. 

Multiply  the  circumference  of  the  base  by  the  altitude  fm 
the  convex  surface.  To  this  add  the  areas  of  the  tioo  ends 
when  the  entire  surface  is  required. 

See  Greometry,  Prop.  1,  B.  X. 

Ex.  1.  What  is  the  convex  surface  of  a  cylinder  whose  alti- 
tude  is  23  feet,  and  the  diameter  of  its  base  3  feet  ? 

Ans.^  216.77  square  feet. 

Ex.  2.  What  is  the  entire  surface  of  a  cylinder  whose  alti- 
tude is  18  feet,  and  the  diameter  of  its  base  5  feet  ? 


Problem  II. 
(120.)  To  find  the  solidity  of  a  cylinder. 

Rule. 
Multiply  the  area  of  the  base  by  the  altitude. 
See  Geometry,  Prop,  2,  B.  X. 

Ex.  1.  What  is  the  solidity  of  a  cylinder  whose  altitude  is 
18  feet  4  inches,  and  the  diameter  of  its  base  2  feet  10  inches  ? 

Ans.^  115.5917  cubic  feet. 
Ex.  2.  What  is  the  solidity  of  a  cylinder  whose  altitude  is 
12  feet  11  inches,  and  the  circumference  of  its  base  5  feet  3 
inches  ? 

Ans.,  28.3308  cubic  feet. 

Problem  III. 
(121.)  To  find  the  surface  of  a  cone. 

Rule. 

Multiply  the  circumference  of  the  base  by  half  the  side  far 
the  convex  surface  ;  to  which  add  the  area  of  the  base  when 
the  entire  surface  is  required. 

See  Greometry.  Prop.  3,  B.  X. 


Mensuration   of   Solids.  86 

Ex.  1.  What  is  the  entire  surface  of  a  cone  whose  side  is  10 
feet  and  the  diameter  of  its  hase  2  feet  3  inches  ? 

Ans.^  39.319  squi  re  feet 
Ex.  2.  "What  is  the  entire  surface  of  a  cone  whose  side  is  15 
feet,  and  the  circumference  of  its  base  8  feet  ? 

Ans.^  65.093  square  feet. 

Problem  IV. 
(122.)  To  find  the  solidity  of  a  cone. 

Rule. 
Multiply  the  area  of  the  hase  by  one  third  of  the  altitude. 
See  Greometry,  Prop.  5,  B.  X. 

•Ex.  1.  What  is  the  solidity  of  a  cone  whose  altitude  is  12 
feet,  and  the  diameter  of  its  base  2 J  feet  ? 

Ans.^  19.635  cubic  feet. 
Ex.  2.  What  is  the  solidity  of  a  cone  whose  altitude  is  25 
feet,  and  the  circumference  of  its  base  6  feet  9  inches  ? 

Ans. 

Problem  "V. 
^123.)  To  find  the  surface  of  a  frustum  of  a  cone. 

*  Rule. 
Multiply  half  the  side  by  the  sum  of  the  circumferences  of 
the  two  bases  for  the  convex  surface;  to  this  add  the  areas 
of  the  two  bases  vjhen  the  entire  surface  is  required. 
See  Greometry,  Prop.  4,  B.  X. 

Ex.  1.  What  is  the  entire  surface  of  a  frustum  of  a  cone, 
the  diameters  of  whose  bases  are  9  feet  and  5  feet,  and  whose 
side  is  16  feet  9  inches  ? 

Ans.,  451.6036  square  feet. 
Ex.  2.  What  is  the  convex  surface  of  a  frustum  of  a  cone 
whose  side  is  10  feet,  and  the  circumferences  of  its  bases  6 
feet  and  4  feet  ? 

.  Ans.,  50  square  feet. 

Problem  VI. 
(124.)  To  find  the  solidity  of  a  frustum  of  a  cone. 


86  Trigonometry. 

Rule 

Add  together  the  areas  of  the  tivo  bases ^  and  a  mean  pro» 
portional  between  them,  and  multiply  the  sum  by  one  third 
of  the  altitude. 

See  Greometry,  Prop.  6,  B.  X. 

If  we  put  R  and  r  for  the  radii  of  the  two  bases,  then  ttR' 
will  represent  the  area  of  one  base,  7rr'  the  area  of  the  other, 
and  irRr  the  mean  proportional  between  them.  Hence,  if  we 
represent  the  height  of  the  frustum  by^i,  its  solidity  will  be 

^(R'+r+Rr). 

Ex.  1.  "What  is  the  solidity  of  a  frustum  of  a  cone  whose 
altitude  is  20  feet,  the  diameter  of  the  greater  end  5  feet,  and 
that  of  the  less  end  2  feet  6  inches  ? 

Ans.,  229.074  cubic  feet 

Ex.  2.  The  length  of  a  mast  is  60  feet,  its  diameter  at  the 
greater  end  is  20  inches,  and  at  the  less  end  12  inches :  what 
is  its  solidity  ?  Ans.,  85.521  cubic  feet 

Problem  Yll. 
(125.)  To  find  the  surface  of  a  sphere. 

Rule. 

Multiply  the  diameter  by  the  circumference  of  a  great  cir- 
cle ;  or.  Multiply  the  square  of  the  diameter  by  3.14159. 

See  Geometry,  Prop.  7,  B.  X. 

Ex.  1.  Required  the  surface  of  the  earth,  its  diameter  be- 
ing 7912  miles.  Ans.,  196,662,896  square  miles. 

Ex.  2.  Required  the  surface  of  the  moon,  its  circumference 
being  6786  miles.  Ans. 

Problem  VIII. 
(126.)  To  find  the  solidity  of  a  sphere. 

Rule. 
Multiply  the  surface  by  one  third  of  the  radius  ;  or,  Mul- 
liply  the  cube  of  the  diameter  by  ^ir ;  that  is,  by  0.5236. 
See  Geometry,  Prop.  8,  B.  X. 
Where  great  accuracy  is  required,  the  value  of  jtt  must  be 


Mensuration   of    Solids. 


87 


taken  to  more  than  four  decimal  places.     Its  value,  correct  to 
ten  decimal  places,  is  .52359,87756. 

Ex.  1.  What  is  the  solidity  of  the  earth,  if  it  be  a  sphere 
7912  miles  in  diameter  ? 

Ans.,  259,332,805,350  cubic  miles. 

Ex.  2.  If  the  diameter  of  the  moon  be  2160  miles,  what  is 
its  solidity  ?  Ans. 

Problem  IX. 
(127.)  To  find  the  surface  of  a  spherical  zone. 

Rule. 
Multiple/  the  altitude  of  the  zone  by  the  circumference  of  a 
^reat  circle  of  the  sphere. 

See  G-eometry,  Prop.  7,  Cor.  1,  B.  X. 

Ex.  1.  If  the  diameter  of  the  earth  be  7912  miles,  what  is 
the  surface  of  the  torrid  zone,  extending  23°  27'  36"  on  each 
side  of  the  equator  ? 

Ans,^  78,293,218  square  miles. 

Let  PEP'Q,  represent  a  meridian  of  the  earth;  EQ,  the 
equator  ;  P,  P'  the  poles  ;  AB  one  of  the 
tropics,  and  GH  one  of  the  polar  circles. 
Then  PK  will  represent  the  height  of  one 
of  the  frigid  zones,  KD  the  height  of  one 
of  the  temperate  zones,  and  CD  half  the 
height  of  the  torrid  zone. 

Each  of  the  angles  ACE,  CAD,  and 
aCK  is  equal  to  23°  27'  36".  p' 

In  the  right-angled  triangle  ACD, 

R  :  AC  :  :  sin.  CAD  :  CD. 
Also,  in  the    ight-angled  triangle  CGrK, 

R  :  CG-  : :  cos.  GCK  :  CK, 
Then  PK=PC-KC. 

Where  great  accuracy  is  required,  the  sine  and  cosine  ol 
23°  27'  36"  must  be  taken  to  more  than  six  decimal  plaoea 
The  following  values  are  correct  to  ten  decimal  places : 
Natural  sine  of  23°  27'  36" =.39810,87431. 
''    cosine  of  23°  27'  36" =.91733,82302. 


88  Trigonometry. 

Ex.  2.  If  the  polar  circle  extends  23°  27'  36"  from  the  pole 
find  the  convex  surface  of  either  frigid  zone. 

Ans.)  8,128,252  square  miles. 
Ex.  3.  On  the  same  suppositions,  find  the  surface  of  each  of 
the  temperate  zones. 

Ans.,  51,056,587  square  miles. 

Problem  X. 
(128.)  To  find  the  solidity  of  a  spherical  segment  with  one 
base. 

Rule. 
Multiply  half  the  height  of  the  segment  by  the  area  of  the 
base,  and  the  cube  of  the  height  by  .5236,  and  add  the  two 
products.  ' 

See  Geometry,  Prop.  9,  B.  X. 

Ex.  1.  What  is  the  solidity  of  either  frigid  zone,  supposing 
the  earth  to  be  7912  miles  in  diameter,  the  polar  circles  ex- 
tending 23°  27'  36"  from  the  poles  ? 

Ans.,  1,292,390,176  cubic  miles. 
(129.)  The  solidity  of  a  spherical  segment  of  two.  bases  is 
the  difference  between  two  spherical  segments,  each  having  a 
single  base. 

Ex.  2.  On  the  same  supposition  as  in  Ex.  1,  find  the  solid- 
ity of  either  temperate  zone. 

•     Ans.,  55,032,766,543  cubic  miles 
Ex.  3.  Find  the  solidity  of  the  torrid  zone. 

Ans.,  146,682,491,911  cubic  miles 

Problem  XI. 
(130.)  To  find  the  area  of  a  spherical  triangle. 

RuJiE. 

Compute  the  surface  of  the  quadrantal  triangle,  or  one 
eighth  of  the  surface  of  the  sphere.  From  the  sum  of  the 
three  angles  subtract  two  right  angles  ;  divide  the  remainder 
by  90,  and  multiply  the  quotient  by  the  quadrantal  triangle 

See  Geometry,  Prop.  20,  B.  IX. 


Mensuration   of    Solids. 


89 


Ex.  1.  What  is  the  area  of  a  triangle  on  a  sphere  whose  di- 
ameter is  10  feet,  if  the  angles  are  dd"",  60°,  and  85°  ? 

Ans.,  8.7266  square  feet. 

Ex.  2.  If  the  angles  of  a  spherical  triangle  measured  on  the 
surface  of  the  earth  are  78°  4'  10",  59°  50'  54",  and  42°  5'  37", 
what  is  the  area  of  the  triangle,  supposing  the  earth  a  sphere, 
of  which  the  diameter  is  7912  miles  ? 

Ans.,  3110.794  square  miles. 

If  the  excess  of  the  angles  above  two  right  angles  is  ex- 
pressed in  seconds,  we  must  divide  it  by  90  degrees  also  ex- 
pressed in  seconds ;  that  is,  by  324,000. 

Problem  XII. 
(131.)  To  find  the  area  of  a  spherical  polygon. 


Rule. 

Compute  the  surface  of  the  quadrantal  triangle.     From 

the  sum  of  all  the  angles  subtract  the  product  of  two  right 

angles  by  the  number  of  sides  less  two  ;  divide  the  remainder, 

by  90,  and  multiply  the  quotient  by  the  quadrantal  triangle. 

See  Greometry,  Prop.  21,  B.  IX. 

Ex.  1.  "What  is  the  area  of  a  spherical  polygon  of  5  sides  on 
a  sphere  whose  diameter  is  10  feet,  supposing  the  sum  of  the 
angles  to  be  640  degrees  ? 

Ans.,  43.633  square  feet 
62°  33'  13' 


Ex.  2.  The  angles  of  a  spherical 
polygon,  measured  on  the  surface 
of  the  earth,  are 


Required  the  area  of  the  polygon. 

Ans.,  5690.477  square  miles 


135°  8' 

26" 

149°  16' 

9" 

111°  45^ 

8" 

105°  59' 

7" 

155°  19' 

12". 

BOOK  IV. 

'    SURVEYING. 

(132.)  The  term  Surveying  includes  the  measurement  ol 
heights  and  distances,  the  determination  of  the  area  of  portions 
01  the  earth's  surface,  and  their  delineation  upon  paper. 

Since  the  earth  is  spherical,  its  surface  is  not  a  plane  sur- 
lace,  and  if  large  portiorfs  of  the  earth  are  to  be  measured,  the 
curvature  must  be  taken  into  account ;  but  in  ordinary  sur- 
veying, the  portions  of  the  earth  are  supposed  to  be  so  small 
that  the  curvature  may  be  neglected.  The  parts  surveyed  are 
therefore  regarded  as  plane  figures. 

(133.)  If  a  plummet  be  fteely  suspended  by  a  line,  and  al- 
lowed to  come  to  a  state  of  rest,  this  line  is  called  a  vertical 
line. 

Every  plane  passing  through  a  vertical  line  is  a  vertica* 
plane. 

A  line  perpendicular  to  a  vertical  line  is  a  horizontal  line. 

A  plane  perpendicular  to  a  vertical  line  is  a  horizontal 
plane. 

A  vertical  angle  is  one  the  plane  of  v^hose  sides  is  vertical. 

A  horizontal  angle  is  one  the  plane  of  vv^hose  sides  is  hori- 
zontal. 

An  angle  of  elevation  is  a  vertical  angle  having  one  side 
horizontal  and  the  other  an  ascending  ^  D 

line,  as  the  angle  BAD. 

An  angle  of  depression  is  a  vortical 
angle  having  one  side  horizontal  and 
the  other  a  descending  line,  as  the  an- 
gle CDA. 

(134.)  When  distances  are  to  be  found  ^ 
by  trigonometrical  computation,  it  is  necessary  to  measure  at 
least  one  line  upon  the  ground,  and  also  as  many  angles  as 
may  be  necessary  to  render  three  parts  of  every  triangle 
known. 


Surveying.  91 

1q  the  measurement  of  lines,  the  unit  commonly  employed 
by  surveyors  is  a  chain  four  rods  or  sixty-six  feet  in  length, 
called  Gunter's  Chain^  from  the  name  of  the  inventor.  This 
chain  is  divided  into  100  links.  Sometimes  a  half  chain  ia 
used,  containing  50  links. 

Hence,         1  chain  ==  100  links   =QQ  feet; 
1  rod     =    25  links   =16|  feet; 
1  link   =7.92  inches  =     f  of  a  foot  nearly. 

(135.)  To  measure  a  horizontal  line. 

To  mark  the  termination  of  the  chain  in  measuring,  ten  iron 
pins  should  be  provided,  about  a  foot  in  length. 

Let  the  person  who  is  to  go  foremost  in  carrying  the  chain, 
and  who  is  called  the  leader,  take  one  end  of  tha  chain  and  the 
ten  pins ;  and  let  another  person  take  the  other  end  of  the 
chain,  and  hold  it  at  the  beginning  of  the  line  to  be  measured. 
When  the  leader  has  advanced  until  the  chain  is  stretched 
tight,  he  must  set  down  one  pin  at  the  end  of  the  chain,  the 
other  person  taking  care  that  the  chain  is  in  the  direction  of 
the  line  to  be  measured.  Then  measure  a  second  chain  in  the 
same  rianner,  and  so  on  until  all  the  marking  pins  are  ex- 
hausted. A  record  should  then  be  made  that  ten  chains  have 
been  measured,  after  which  the  marking  pins  should  be  re- 
turned to  the  leader,  and  the  measurement  continued  as  be- 
fore until  the  whole  line  has  been  passed  over. 

It  is  generally  agreed  to  refer  all  surfaces  to  a  horizontal 
planfe.  Hence,  when  an  inclined  surface,  like  the  side  of  a 
hill,  is  to  be  measured,  the  chain  should  be  maintained  in  a 
horizontal  position.  For  this  purpose,  in  ascending  a  hill,  the 
hind  end  of  the  chain  should  be  raised  from  the  ground  until 
it  is  on  a  level  with  the  fore  end,  and  should  be  held  vertically 
over  the  termination  of  the  preceding  chain.  In  descending  a 
hill,  the  fore  end  of  the  chain  should  be  raised  in  the  same 
manner. 


INSTRUMENTS  FOR  MEASURING  ANGLES. 

In  measuring  angles,  some  instrument  is  used  which  con- 
tains a  portion  of  a  graduated  circle  divided  into  degrees  and 
minutes.     These  instruments  may  be  adapted  to  measuring 


.92 


Trigonometry. 


either  horizontal  or  vertical  angles.     The  instrument  most  fre- 
quently employed  for  measuring  horizontal  angles  is  called 

The  Surveyor's  Compass. 
»  (136.)  The  piincipal  parts  of  this  instrument  are  a  compass- 
box,  a  magnetic  needle,  two  sights,  and  a  stand  for  its  support. 
The  compass-box,  ABC,  is  circular,  generally  about  six  inches 
in  diameter,  and  at  its  center  is  a  small  pin  on  which  the  mag- 
netic needle  is  balanced.  The  circumference  of  the  box  is  di- 
vided into  degrees,  and  sometimes  to  half  degrees  ;  and  the  de- 
grees are  numbered  from  the  extremities  of  a  diameter  both 
ways  to  90°.     The  sights,  DE,  FGr,  are  placed  at  right  angles 


lo  the  plane  of  the  graduated  circle,  and  in  each  of  these  there 
is  a  large  and  small  aperture  for  convenience  of  observation 
The  instrument,  when  used,  is  mounted  on  a  tripod,  or  a  single 
staff  pointed  with  iron  at  the  bottom,  so  that  it  may  be  firmly 
placed  in  the  ground. 

Sometimes  two  spirit  levels,  H  and  K,  are  attached,  to  indi- 
cate when  the  plane  of  the  graduated  circle  is  brought  into  a 
horizontal  position. 

(137.)  When  the  magnetic  needle  is  supported  so  as  to  thrn 
freely,  and  is  allowed  to  come  to  a  state  of  rest,  the  direction 
it  assumes  is  called  the  magnetic  meridian,  one  end  of  the 
needle  indicating  the  north  point  and  the  other  the  south. 

A  horizontal  lino,  perpendicular  to  a  meridian  is  an  east  and 
west  line 


Surveying. 


93 


N 


All  the  meridians  passing  through,  a  survey  of  moderate  ex- 
tent, are  considered  as  straight  lines  parallel  to  each  other. 

The  bearing  or  course  of  a  line  is  the  angle  which  it  makes 
with  a  meridian  passing  through  one  end ;  and  it  is  reckoned 
from  the  north  or  south  point  of  the  horizon,  toward  the  east 
or  west. 

Thus,  if  NS  represent  a  meridian,  and  the  angle  NAB  is  40'', 
then  the  hearing  of  AB  from  the  point  A  is 
40°  to  the  west  of  north,  and  is  written  N.  40° 
W.,  and  read  north  forty  degrees  west. 

The  reverse  bearing  of  a  line  is  the  hearing 
taken  from  the  other  end  of  the  line. 

The  forward  hearing  and  reverse  hearing 
of  a  line  ar^  equal  angle?,  hut  lie  between  di- 
rectly opposite  points.  Thus,  if  the  hearing 
of  AB  from  A  is  N.  40°  W.,  the  hearing  of  the 
same  line  from  B  is  S.  40°  E. 

(138.)  For  measuring  vertical  angles,  the  instrument  com. 
monly  used  is 


A  Quadrant. 

It  consists  of  a  quarter  of  a  circle,  usually  made  of  brass, 
and  its  limb,  AB,  is  divided  into 
degrees  and  minutes,  numbered 
from  A  up  to  90°.  It  is  furnish- 
ed either  with  a  pair  of  plain 
sights  or  witli  a  telescope,  CD, 
which  is  to  be  directed  toward 
the  object  observed.  A  plumb 
line,  CE,  is  suspended  from  the 
center  of  the  quadrant,  and.  in- 
dicates when  the  radius  CB  is 
brought  into  a  vertical  position. 

To  measure  the  angle  of  elevation,  for  example,  of  the  top 
of  a  tower,  point  the  telescope,  CD,  toward  the  tower,  keeping 
the  radius,  CB,  in  a  vertical  position  by  means  of  the  plumb 
line,  CE.  Move  the  telescope  until  the  given  object  is  seen  in 
the  middle  of  the  field  of  view.  The  center  of  the  field  is  in- 
dicated by  two  wires  placed  in  the  focus  of  the  object-glass  of 


94 


Trigonometry. 


the  telescope,  one  wire  being  vertical  and  the  other  horizontal. 
When  the  horizontal  wire  is  made  to  coiacide  with  the  sum- 
mit of  the  tower,  the  angle  of  elevation  is  shown  upon  the  arc 
AuB  by  means  of  an  index  w^hich  moves  with  the  telescope. 

As  the  arc  is  not  commonly  divided  into  parts  smaller  than 
half  degrees,  when  great  accuracy  is  required,  some  contriv- 
ance is  needed  for  obtaining  smaller  fractions  of  a  degree. 
This  is  usually  effected  by  a  vernier. 

(139.)  A  Vernier  is  a  scale  of  small  extent,  graduated  in 
such  a  manner  that,  being  moved  by  the  side  of  a  fixed  scale, 
we  are  enabled  to  measure  minute  portions  of  this  scale.'  The 
length  of  this  movable  scale  is  equal  to  a  certain  number  of 
parts  of  that  to  be  subdivided,  but  it  is  divided  into  parts  one 
more  or  one  less  than  those  of  the  primary  scale  taken  for  the 
length  of  the  vernier.  Thus,  if  we  wish  to  measure  hundredths 
of  an  inch,  as  in  the  case  of  a  barometer,  we  first  divide  an 
inch  into  ten  equal  parts.  We  then  construct  a  vernier  equal 
in  length  to  11  of  these  divisions,  but  divide  it  into  10  equal 
parts,  by  which  jneans  each  division  on  the  vernier  is  -^-^ik 
longer  than  a  division  of  the  primary  scale. 

Thus,  let  AB  be  the  upper  end  of  a  barometer  tube,  the  mer- 
cury standing  at  the  point  C ;  the  scale  is 
divided  into  inches  and  tenths  of  an  inch, 
and  the  middle  piece,  numbered  from  1 
to  9,  is  the  vernier  that  slides  up  and 
down,  having  10  of  its  divisions  equal  to 
11  divisions  of  the  scale,  that  is,  to  l^ths 
of  an  inch.  Therefore,  each  division  of 
the  vernier  ts  yVo^hs  of  an  inch ;  or  one 
division  of  the  vernier  exceeds  one  divi- 
sion of  the  scale  by  yloth  of  an  inch. 
Now,  as  the  sixth  division  of  the  vernier 
(in  the  figure)  coincides  with  a  division 
of  the  scale,  the  fifth  division  of  the  ver- 
nier will  stand  7^0 th  of  an  inch  above  the  nearest  division  of 
the  scale ;  the  fourth  division  ylo^hs  of  an  inch,  and  the  top 
of  the  vernier  will  be  yf  nths  of  an  inch  above  the  next  lower 
division  of  the  scale  ;  i.  e.,  the  top  of  the  vernier  coincides  with 
29  G6  inches  upon  the  scale.     In  practice,  therefore,  we  ob- 


Surveying.  95 

serve  what  division  of  the  vernier  coincides  with  a  division  of 
the  scale ;  this  will  show  the  hundredths  of  an  inch  to  be  added 
to  the  tenths  next  below  the  vernier  at  the  top. 

A  similar  contrivance  is  applied  to  graduated  circles,  to  ob- 
tain the  value  of  an  arc  with  greater  accuracy.  If  a  circle  is 
graduated  to  half  degrees,  or  30',  and  we  wish  to  measure  sin- 
gle minutes  by  the  vernier,  we  take  an  arc  equal  to  31  divi- 
riions  upon  the  limb,  and  divide  it  into  30  equal  parts.  Then 
each  division  of  the  vernier  will  be  equal  to  ^ '  ths  of  a  degree, 
while  each  division  of  the  scale  is  ||ths  of  a  degree.  That  is, 
each  space  on  the  vernier  exceeids  one  on  the  limb  by  1'. 

In  order,  therefore,  to  read  an  angle  for  any  position  of  the 
vernier,  we  pass  along  the  vernier  until  a  line  is  found  coin- 
ciding with  a  line  of  the  limb.  The  number  of  this  line  from 
the  zero  point  indicates  the  minutes  which  are  to  be  added  to 
the  degrees  and  half  degrees  taken  from  the  graduated  circle. 
Sometimes  a  vernier  is  attached  to  the  common  surveyor'^ 
compass. 

(140.)  An  instrument  in  common  use  for  measuring  both 
horizontal  and  vertical  angles  is 

The  Theodolite. 

The  theodolite  has  two  circular  brass  plates,  C  and  D  (see  fig. 
next  page),  the  former  of  which  is  called  the  vernier  plate,  and 
the  latter  the  graduated  limb.  Both  have  a  horizontal  motion 
about  the  vertical  axis,  E.  This  axis  consists  of  two  parts,  one 
external,  and  the  other  internal ;  the  former  secured  to  the- 
graduated  limb,  D,  and  the  latter  to  the  vernier  plate,  C,  so  that 
the  vernier  plate  turns  freely  upon  the  lower.  The  edge  of  the 
lov/er  plate  is  divided  into  degrees  and  half  degrees,  and  this 
is  subdivided  by  a  vernier  attached  to  the  upper  plate  into 
single  minutes.     The  degrees  are  numbered  from  0  to  360. 

The  parallel  plates,  A  and  B,  are  held  together  by  a  ball 
which  rests  in  a  socket.  Four  screws,  three  of  which,  a,  a,  a, 
are  shown  in  the  figure,  turn  in  sockets  fixed  to  the  lower  plate, 
while  their  heads  press  against  the  under  side  of  the  upper 
plate,  by  which  means  the  instrument  is  leveled  for  observa- 
tim.  The  whole  ^ests  upon  a  tripod,  which  is  firmly  attached 
to  the  body  of  the  instrument. 


96 


^Vo^mT^ 


To  the  vernier  plate,  two  spirit-levels,  c,  c,  are  attached  at 
right  angles  to  each  other,  to  determine  when  the  graduated 
limb  is  horizontal.  A  compass,  also,  is  placed  at  F.  Two 
frames,  one  of  which  is  seen  at  N,  support  the  pivots  of  the 
horizontal  axis  of  the  vertical  semicircle  KL,  on  which  the  tel- 
escope, GrH,  is  placed.  One  side  of  the  vertical  arc  is  divided 
into  degrees  and  half  degrees,  and  it  is  divided  into  single  min- 
utes by  the  aid  of  its  vernier.  The  graduation  commences  at 
the  middle  of  the  arc,  and  reads  both  ways  to  90°.  Under  and 
parallel  to  the  telescope  is  a  spirit-level,  M,  to  show  when  the 
telescope  is  brought  to  a  horizontal  position.  To  enable  us  to 
direct  the  telescope  upon  an  object  with  precision,  two  lines 
called  wires  are  fixed  at  right  angles  to  each  other  in  the  focus 
of  the  telescope. 


To  measure  a  Horizontal  Angle  with  the  Theodolite. 
(141.)  Place  the  instrument  exactly  over  the  station  from 
which  the  angle  is  to  be  measured ;  then  level  the  instrument 
by  means  of  the  screws,  a,  a,  bringing  the  telescope  over  each 
pair  alternately  until  the  two  spirit-levels  on  the  vernier  plate 
retain  their  position,  while  the  instrument  is  turned  entirely 
round  upon  its  axis.     Direct  the  telescope  to  one  of  the  objects 


Surveying.  97 

to  be  observed,  moving  it  until  the  cross-wires  and  object  co- 
incide. Now  read  off  the  degrees  upon  the  graduated  limb, 
and  the  minutes  indicated  by  the  vernier.  Next,  release  the 
upper  plate  (leaving  the  graduated  limb  undisturbed),  and 
move  it  round  until  the  telescope  is  directed  to  the  second  ob- 
ject, and  make  the  cross-wires  bisect  this  object,  as  was  done 
by  the  first.  Again,  read  off  the  vernier ;  the  difference  be- 
tween this  and  the  former  reading  will  be  the  angle  required. 
The  magnetic  bearing  of  an  object  is  determined  by  simply 
reading  the  angle  pointed  out  by  the  compass-needle  when  the 
object  is  bisected. 

To  measure  an  Angle  of  Elevation  with  the  Theodolite, 
(142.)  Direct  the  telescope  toward  the  given  object  so  that 
it  may  be  bisected  by  the  horizontal  wire,  and  then  read  off 
the  arc. upon  the  vertical  semicircle.  After  observing  the  ob- 
ject with  the  telescope  in  its  natural  position,  it  is  well  to  re- 
volve the  telescope  in  its  supports  until  the  level  comes  upper- 
most, and  repeat  the  observation.  The  mean  of  the  two  meas- 
ures may  be  taken  as  the  angle  of  elevation. 

By  the  aid  of  the  instruments  now  described,  we  may  de- 
termine the  distance  of  an  inaccessible  object,  and  its  height 
abov5  the  surface  of  the  earth. 

HEIGHTS  AND  DISTANCES. 
Problem  I. 

(143.)  To  determine  the  height  of  a  vertical  object  situatea 
on  a  horizontal  plane. 

Measure  from  the  object  to  any  convenient  distance  in  a 
straight  line,  and  then  take  the  angle  of  elevation  subtended 
by  the  object. 

If  we  measure  the  distance  DE,  and 
the  angle  of  elevation  CDE,  there  will 
be  given,  in  the  right-angled  triangle 
CDE,  the  base  and  the  angles,  to  find 
the  perpendicular  CE  (Art.  46).  To 
this  we  must  add  the  height  of  the  in- 

D  '' 
strument,  to  obtain  the  entire  height      ~ 

of  the  object  alove  the  plane  AB. 

a 


/ 

/     J 

u 

1 

1 
1 

1 

98  Trigonometry 

Ex.  1.  Having  measured  AB  equal  to  100  feet  from  tho 
bottom  of  a  tower  on  a  horizontal 
plane,  I  found  the  angle  of  elevation, 
CDE,  of  the  top  to  be  47°  30',  the 
center  of  the  quadrant  being  five 
feet  above  the  ground.  What  is  the 
height  of  the  tower  ? 
R  :  tang.  CDE  : :  DE  :  CE  =  109.13. 
To  which  add  five  feet,  and  we  obtain 
the  height  of  the  tower,  114.13  feet.     ^  ^ 

Ex.  2.  From  the  edge  of  a  ditch  18  feet  wide,  surrounding 
a  fort,  the  angle  of  elevation  of  the  wall  was  found  to  be  62° 
40'.  Required  the  height  of  the  wall,  and  the  length  of  a  lad- 
der necessary  to  reach  from  my  station  to  the  top  of  it. 

Ans.  The  height  is  34.82  feet.     Length  of  ladder,  39.20  feet. 

Problem  II. 

(144.)  To  find  the  distance  of  a  vertical  object  whose  height 
is  known. 

Measure  the  angle  of  elevation,  and  we  shall  have  given  the 
angles  and  perpendicular  of  a  right-angled  triangle  to  find  the 
base  (Art.  46). 

Ex.  1.  The  angle  of  elevation  of  the  top  of  a  tower  whose 
height  was  known  to  be  143  feet,  was  p 

found  to  be  35°.     What  was  its  dis-     ^ 
tance  ?    . 

Here  we  have  given  the  angles  of  the 
triangle  ABC,  and  the  side  CB,  to  find 
AB. 

Ans.,  204.22  feet. 

If  the  observer  were  stationed  at  the  ^ 
top  of  the  tower  BC,  he  might  find  the  length  of  the  base  AB 
by  measuring  the  angle  of  depression  DC  A,  which  is  equal  to 
BAC. 

Ex.  2.  From  the  top  of  a  ship's  mast,  which  was  80  feet 
above  the  water,  the  angle  of  depression  of  another  ship's  hull 
was  found  to  be  20°.     What  was  its  distance  ? 

Ans.,  219.80  feet 


Surveying. 


99 


Problem  III. 

(145.)  To  find  the  height  of  a  vertical  object  standing  on 
tLU  inclined  plane. 

Measure  the  distance  from  the  object  to  any  convenient  sta- 
tion, and  observe  the  angles  which  the  base-line  makes  with 
lines  drawn  from  its  two  ends  to  the  top  of  the  object. 

If  we  measure  the  base-line  AB,  and  the  two  angles  ABC, 
BAG,  then,  in  the  triangle  ABC, 
we  shall  have  given  one  side  and 
the  angles  to  find  BC. 

Ex.  1.  Wanting  to  know  the 
height  of  a  tower  standing  on  an 
inclined  plane,  BD,  I  measured 
from  the  bottom  of  the  tower  a 
distance,  AB,  equal  to  165  feet ; 
also  the  angle  ABC,  equal  to 
107°  18',  and  the  angle  BAC, 
equal  to  33°  35'.  Required  the 
height  of  the  object. 

sin.  ACB  :  AB  :  :  sin.  BAC  :  BC= 144.66  feet. 

The  height,  BC,  may  also  be  found  by  measuring  the  di^ 
tances  BA,  AD,  and  taking  the  angles  BAC,  BDC.  The  dif- 
ference between  the  anorles  BAC  and  BDC  will  be  the  angle 


ACD.     There  will  then  be 


■to 

given,  in  the  triangle  DAC,  one 
side  and  all  the  angles  to  find  AC  ;  after  which  we  shall  have, 
in  the  triangle  ABC,  two  sides  and  the  included  angle  to  find 
l^C. 

Ex.  2.  A  tower  standing  on  the  top  of  a  declivity,  I  meas- 
ured 75  feet  from  its  base,  and  then  took  the  angle  BAC,  47° 
50' ;  going  on  in  the  same  direction  40  feet  further,  I  took  the 
angle  BDC,  38°  30'.     What  was  the  height  of  the  tower  ? 

Ans.,  117.21  feet. 


Problem  IY. 
(146.)  To  find  the  distance  of  an  inaccessible  object. 
Measure  a  horizontal  base-line,  and  also  the  angles  between 
tliis  line  and  lines  drawn  from  each  station  to  the  object.     Let 
C  be  the  object  inaccessible  from  A  and  B.     Then,  if  the  dis- 


100 


TRiGONOMETRY. 


tanee  between  the  stations  A  and  B  be  measured,  as  also  the 
angles  at  A  and  B,  there  will  be  given,  in 
the  triangle  ABC,  the  side  AB  and  the  an- 
gles, to  find  AC  and  BC,  the  distances  of  the 
object  from  the  two  stations. 

Ex.  1.  Being  on  the  side  of  a  river,  and 
wanting  to  know  the  distance  to  a  house 
which  stood  on  the  other  side,  I  measured    A     '  B 

400  yards  in  a  right  line  by  the  side  of  the  river,  and  found 
that  the  two  angles  at  the  ends  of  this  line,  formed  by  the 
other  end  and  the  house,  were  73°  15'  and  68°  2'.  What  was 
the  distance  between  each  station  and  the  house  ? 

The  angle  C  is  found  to  be  38°  43'.     Then 

A:  BC=612.38; 
B  :  AC-593.09. 

Ex.  2.  Two  ships  of  war,  wishing  to  ascertain  their  distance 
from  a  fort,  sail  from  each  other  a  distance  of  half  a  mile,  when 
they  find  that  the  angl*  formed  between  a  line  from  one  to 
the  other,  and  from  each  to  the  fort,  are  85°  15'  and  83°  45'. 
What  are  the  respective  distances  from  the  fort  ? 

Ans.,  4584.52  and  4596.10  yards. 


sin.  C  :  AB 


' '  (  sin. 


Problem  V. 

(147.)  To  find  the  distance  beivjeen  two  objects  separated 
by  an  impassable  barrier. 

Measure  the  distance  from  any  convenient  station  to  each 
of  the  objects,  and  the  angle  included  between  those  lines. 

If  we  wish  to  know  the  distance  between  the  places  C  and 
B,  both  of  which  are  accessible,  but  sep- 
arated from  each  other  by  water,  we  may 
measure  the  lines  AC  and  AB,  and  also 
the  angle  A.  We  shall  then  have  given 
two  sides  of  a  triangle  and  the  included 
angle  to  find  the  third  side. 

Ex.  1.  The  passage  between  the  two 
objects  C  and  B  b&ing  obstructed,  I  measured  from  A  to  C  735 
rods,  and  from  A  to  B  840  rods ;  also,  the  angle  A,  equal  to 
55°  40'.     What  is  the  distance  of  the  places  C  and  B  ? 

Ans.,  741.21  rods 


Surveying.  101 

Ex.  2.  In  order  to  find  the  distance  between  two  objects,  C 

-d  B,  which  could  not  be  directly  measured,  I  measured 

\  om  C  to  A  652  yards,  and  from  B  to  A  756  yards ;  also,  the 

angle  A  equal  to  142°  25'.     What  is  the  distance  between  the 

objects  C  and  B? 


Problem  YI. 

(148.)  To  find  the  height  of  an  inaccessible  object  above  a 
horizontal  plane. 

First  Method. — Take  two  stations  in  a  vertical  plane  pass- 
ing through  the  top  of  the  object ;  measure  the  distance  be- 
tween the  stations  and  the  angle  of  elevation  at  each. 

If  we  measure  the  base  AB,  and  the  angles  DAC,  DBC, 
then,  since  CBA  is  the  sup-  C 

plement  of  DBC,  we  shall  ^----^^^^Imv 

have,  in  the  triangle  ABC,  ^^.^-^'^'^^^^^^'^f^^ 

one  side  and  all  the  angles       ^^^^^"^^"^  "B^^i^^.^^^Mw^^^ 
to  find  BC.     Then,  in  the     "^^^^^'       '"        1) 
right-angled  triangle  DBC,  we  shall  have  the  hypothenuse  and 
the  angles  to  find  DC. 

Ex.  1.  What  is  the  perpendicular  height  of  a  hill  whose  an- 
gle of  elevation,  taken  at  the  bottom  of  it,  was  46° ;  and  100 
yards  farther  off,  on  a  level  with  the  bottom  of  it,  the  angle 
was  31°  ? 

Ans.^  143.14  yards. 

Ex.  2.  The  angle  of  elevation  of  a  spire  I  found  to  be  58°, 
and  going  100  yards  directly  from  it,  found  the  angle  to  be 
only  32°.  What  is  the  height  of  the  spire,  supposing  the  in- 
strument to  have  been  five  feet  above  the  ground  at  each  ob- 
servation ? 

Ans.^  104.18  yards. 

(149.)  Second  Method. — Measure  any  convenient  base-line, 
also  the  angles  between  this  base  and  lines  drawn  from  each 
of  its  extremities  to  the  foot  of  the  object,  and  the  angle  of 
elevation  at  one  of  the  stations. 

Let  DC  be  the  given  object.  If  we  measure  the  horizontal 
base-line  AB,  and  the  angles  CAB,  CBA,  we  can  compute  the 
distance  BC.     Also,  if  we  observe  the  angle  of  elevation  CBD, 


:^^'^^^ 


102 


Trigonometry. 


we  shall  have  given,  in  the  right-angled  triangle  BCD,  the 
base  and  angles  to  find  the  perpen- 
dicular. 

Ex.  1.  Being  on  one  side  of  a  river, 
and  wanting  to  know  the  height  of  a 
spire  on  the  other  side,,  I  measured 
500  yards,  AB,  along  the  side  of  the 
river,  and  found  the  angle  ABC =74° 
14',  and  BAG =49°  23' ;  also,  the  an- 
gle  of  elevation  CBD  =  11°  15'.  Required  the  height  of  the 
spire.  Ans.,  271.97  feet. 

Ex.  2.  To  find  the  height  of  an  inaccessible  castle,  I  meas- 
ured a  line  of  73  yards,  ^nd  at  each  end  of  it  took  the  angle  of 
position  of  the  object  and  the  other  end,  and  found  the  one  to  be 
90°,  and  the  other  61°  45' ;  also,  the  elevation  of  the  castle  from 
the  latter  station,  10°  35'.     Required  the  height  of  the  castle. 

Ans.,  86.45  feet. 


Problem  VII. 

(150.)  To  find  the  distance  betvjeen  two  inaccessible  objects. 

Measure  any  convenient  base-line,  and  the  angles  between 
this  base  and  lines  drawn  from  each  of  its  extremities  to  each 
of  the  objects. 

Let  C  and  D  be  the  two  inaccessible  objects.  If  we  meas- 
ure  a  base-line,  AB,  and  the  an- 
gles DAB,  DBA,  CAB,  CBA, 
then,  in  the  triangle  DAB,  we 
shall  have  given  the  side  AB 
and  all  the  angles  to  find  BD ; 
also,  in  the  triangle  ABC,  we 
shall  have  one  side  and  all  the 
angles  to  find  BC ;  and  then,  in  -^  ^ 

the  triangle  BCD,  we  shall  have  two  sides,  BD,  BC,  with  the 
included  angle,  to  find  DC.  # 

Ex.  1.  Wanting  to  know  the  distance  between  a  house  and 
a  mill,  which  were  separated  from  me  by  a  river,  I  measured 
a  base-line,  AB,  300  yards,  and  found  the  angle  CAB =58° 
20',  CAD=37°,  ABD=53°  30',  DBC=45°  15'.  What  is  the 
distance  of  the  house  from  the  mill  ?       Ans.,  479.80  yards. 


Surveying.  103 

Ex.  2.  Wanting  to  know  the  distance  between  two  inaccessi 
ble  objects,  C  and  D,  I  nieasured  a  base-line,  AB,  28.76  rods, 
and  found  the  angle  CAB=33%  CAD=66°,  DBA=59°  45', 
and  DBC=76°.     What  is  the  distance  from  C  to  D  ? 

Ans.,  97 '696  rods. 

THE  DETERMINATION  OF  AREAS. 

(151.)  The  area  or  content  of  a  tract  of  land  is  the  horizon- 
tal surface  included  within  its  boundaries. 

When  the  surface  of  the  ground  is  broken  and  uneven,  it  is 
very  difficult  to  ascertain  exactly  its  actual  surface.  Hence 
it  has  been  agreed  to  refer  every  surface  to  a  horizontal  plane  ; 
and  for  this  reason,  in  measuring  the  boundary  lines,  it  is  nec- 
essary to  reduce  them  all  to  horizontal  lines. 

The  measuring  unit  of  surfaces  chiefly  employed  by  survey- 
ors is  the  acre  J  or  ten  square  chains. 

One  quarter  of  an  acre  is  called  a  rood. 

Since  a  chain  is  four  rods  in  length,  a  square  chain  containsi 
sixteen  square  ,rods  ;  and  an  acre,  or  ten  square  chains,  con- 
tains 160  square  rods.  Square  rods  are  called  perches.  The 
area  of  a  field  is  usually  expressed  in  acres,  roods,  and  perches, 
designated  by  the  letters  A.,  R.,  P. 

When  the  lengths  of  the  bounding  lines  of  a  field  are  given 
in  chains  and  links,  the  area  is  obtained  in  square  chains  and 
square  links.  Now,  since  a  link  is  jj^  of  a  chain,  a  square 
link  will  be  tfoXtIo  of  a  square  chain;  that  is,  voioo  of  a 
chain.     Hence  we  have  the  following 

Table. 

1  square  chain=  10,000  square  links. 

1  acre=10  square  chains =100,000  square  links. 

1  aore=4  roods=160  perches. 
If,  then,  the  linear  dimensions  are  links,  the  area  will  be  ex- 
pressed in  square  links,  and  may  be  reduced  to  square  chains 
by  cutting  o^  four  places  of  decimals  ;  iifive  places  be  cut  off, 
the  remaining  figures  will  be  acres.  If  the  decimal  part  of  an 
acre  be  multiplied  by  4,  it  will  give  the  roods,  and  the  result- 
ing decimal,  multiplied  by  40,  will  give  the  perches. 


104  Trigonometry. 

(152.)  The  difference  of  latitude^  or  the  northing  or  vouth* 
ing  of  a  line,  is  the  distance  that  one  end  is  further  north  or 
south  than  the  other  end. 

Thus,  if  NS  be  a  meridian  passing  through  the  end  A  of  the 
line  AB,  and  BC  be  perpendicular  to  NS,  then  is 
AC  the  difference  of  latitude,  or  northing  of  AB. 

The  departure^  or  the  easting  or  ivesting  of  a 
*ine,  is  the  distance  that  one  end.  is  further  east  or 
west  than  the  other  end. 

Thus  BC  is  the  departure  or  westing  of  tl^e  line 
AB. 

It  is  evident  that  the  distance,  difference  of  lat- 
itude, and  departure  form  a  right-angled  triangle, 
of  which  the  distance  is  the  hypothenuse. 

The  meridian  distance  of  a  point  is  the  perpendicular  let  fall 
from  the  given  point  on  some  assumed  meridian,  and  is  east  or 
west  according  as  this  point  lies  on  the  east  or  west  side  of  the 
meridian. 

The  meridian  distance  of  a  line  is  the  distance  of  the  middle 
point  of  that  line  from  some  assumed  meridian. 

(153.)  When  a  piece  of  ground  is  to  be  surveyed,  we  begin 
at  one  corner  of  the  field,  and  go  entirely  around  the  field, 
measuring  the  length  of  each  of  the  sides  with  a  chain,  and 
their  bearings  with  a  compass. 

Plotting  a  Survey. 
When  a  field  has  been  surveyed,  it  is  easy  to  draw  a  plan 
of  it  on  paper.  For  this  purpose,  draw  a  line  to  represent  the 
meridian  passing  through  the  first  station  ;  then  lay  off  an  an- 
gle equal  to  the  angle  which  the  first  side  of  the  field  makes 
with  the  meridian,  and  take  the  length  of  the  side  from  a  scale 
of  equal  parts.  Through  the  extremity  of  this  side  draw  a 
second  meridian  parallel  to  the  first,  and  proceed  in  the  same 
manner  with  the  remaining  sides.  This  method  will  be  easily 
\inderstood  from  an  example. 

Example  1.  .  ' 

Draw  a  plan  of  a  field  from  the  following  courses  and  dis« 
tances,  as  given  in  the  field-book. 


Surveying. 


105 


Stations, 

Bearings. 

Distances. 

1 

N.  45°  E. 

9.30  chains. 

2 

a  60°  E. 

11.85       " 

3 

S.  20°  W. 

5.30       ^' 

4 

S.  70°  W. 

10.90       " 

5 

N.  31°  W. 

9.40      " 

Draw  NS  to  represent  a  meridian  line ;  in  NS  take  any  con- 
venient point,  as  A,  for  the  first  station,  and  lay  off  an  angle, 
NAB,  equal  to  45°,  the  bear- 
ing from  A  to  B,  which  will 
give  the  direction  from  A  to 
B.  Then,  from  the  scale  of 
equal  parts,  make  AB  equal 
to  9.30,  the  length  of  the  first  a 
side ;  this  will  give  the  sta- 
tion B.  Through  B  draw  a 
second  meridian  parallel  to 
NS  ;  lay  off  an  angle  of  60°, 
and  make  the  line  BC  equal  to 
11.85.  Proceed  in  the  same 
manner  with  the  other  sides.  If  the  survey  is  correct,  and  the 
plotting  accurately  performed,  the  end  of  the  last  side,  EA, 
will  fall  on  A,  the  place  of  beginning.  This  plot  is  made  on  a 
scale  of  10  chains  to  an  inch. 

(154.)  To  avoid  the  inconvenience  of  drawing  a  meridian 
through  each  angle  of  the  field,  the  sides  may  be  laid  down 
from  the  angles  which  they  make  with  each  other,  instead  of 
the  angles  which  they  make  with  the  meridian.  Reverse  one 
of  the  bearings,  if  necessary,  so  that  both  bearings  may  run 
from  the  same  angular  point ;  then  the  angle  which  any  two 
contiguous  sides  makef»with  each  other  may  be  determined 
from  the  following 

Rules. 

1.  If  both  courses  are  north  or  south,  and  both  east  or  west, 
subtract  the  less  fro7n  the  greater. 

2.  If  both  are  north  or  south,  but  one  east  and  the  other 
west,  add  them  together. 


106 


Trigonometry. 


3.  If  one  is  north  and  the  other  south,  but  both  east  or  west, 
subtract  their  sum  from  180°. 

4.  If  one  is  north  and  the  other  south,  one  east  and  the  other 
west,  subtract  their  difference  from  180°. 

Thus  the  angle  CAB  is  equal  to 
i^AB-NAC. 

The  angle  CAD  is  equal  to  NAG 
hNAD. 

The  angle  DAF  is  equal  to  180° 
^(NAD-hSAF).  ' 

The  angle  CAF  is  equal  to  180° 
-(SAF-NAC). 

In  the  preceding  example  we  ac- 
cordingly find  the  angle 

ABC=105°.  DEA=101°. 

BCD=100°.  EAB=104°. 

CDE  =  130°. 
With  these  angles  the  field  may  be  plotted  without  drawing 
parallels. 


Example  2. 
The  following  field  notes  are  given  to  protract  the  survey 


Stations. 

Bearings. 

Distances. 

1 

N.  50°  30'  E. 

16.50  chains. 

2 

S.  68°  15'  E. 

14.20       *' 

3 

S.     9°  45'  E. 

8.45       " 

4 

S.  21°    0'  W. 

6.84       ^' 

5 

S.  73°  30'  W. 

12.31       '' 

6 

N.  78°  15'  W. 

9.76       " 

7 

N.  15°  30'  W. 

11.55       " 

The  Traverse  Table. 
(155.)  The  accompanying  traverse  table  shows  the  difference 
of  latitude  and  the  departure  to*four  decimal  places,  for  dis- 
tances from  1  to  10,  and  for  bearings  from  0°  to  90°,  at  inter- 
vals of  15'.  If  the  bearing  is  less  than  45°,  the  angle  will  be 
found  on  the  left  margin  of  one  of  the  pages  of  the  table,  dnd 
the  distance  at  the  top  or  bottom  of  the  page ;  the  difference 


SURVKYING.  107 

of  latitude  will  be  found  in  the  column  headed  Lat,  at  the  top 
of  the  page,  and  the  departure  in  the  column  headed  Dep.  If 
the  bearing  is  more  than  45°,  the  angle  will  be  found  on  the 
right  margin,  and  the  difference  of  latitude  will  be  found  in 
the  column  marked  Lat.  at  the  bottom  of  the  page,  and  the 
departure  in  the  other  column.  The  latitudes  and  departures 
for  different  distances  with  the  same  bearing  are  proportional 
to  the  distances.  Therefore  the  distances  may  be  reckoned  as 
tens,  hundreds,  or  thousands,  if  the  place  of  the  decimal  point 
in  each  departure  and  difference  of  latitude  be  changed  ac- 
cordingly. 

Ex.  1.  To  find  the  latitude  and  departure  for  the  course  45° 
and  the  distance  93. 

Under  distance  9  on  page  141,  and  opposite  45°,  will  be 
found  latitude  6.3640  and  departure  6.3640.  Hence,  for  dis- 
tance 90,  the  latitude  is  63.640,  and  adding  the  latitude  for  the 
distance  3,  viz.,  2.121,  we  find  the  latitude  for  distance  93  to 
be  65.761. 

Ex.  2.  To  find  the  latitude  and  departure  for  the  course  60* 
and  the  distance  11.85. 


The  latitude  for 

10  is  5.0000. 

Departure  for  10  is 

8.6603. 

u            u            a 

1  is    .5000. 

"          "      1  is 

.8660. 

a           a            (( 

.8  is    .4000. 

"          ''     .8  is 

.6928. 

a           u            a 

.05  is    .0250. 

"          "  .05  is 

.0433. 

Latitude  for  11.85  is  5.9250. 


Depart,  for  11.85  is  10.2624. 


Ex.  3.  To  find  the  latitude  and  departure  for  the  course 
20°  and  the  distance  5.30. 

Ans.  Latitude  4.98,  and  departure  1.81. 

The  traverse  table  may  be  used  not  only  for  obtaining  de- 
parture and  difference  of  latitude,  but  for  finding  by  inspection 
the  sides  and  angles  of  any  right-angled  triangle  ;  for  the  lati- 
tude and  departure  form  the  two  legs  of  a  right-angled  trian- 
gle, of  w  hich  the  distance  is  the  hypothenuse,  and  the  courso 
is  one  of  the  acute  angles.     ' 

In  this  manner  we  find  the  latitude  and  departure  for  each 
side  of  the  field  given  in  Example  1,  page  105,  to  be  as  in  the 
following  table : 


108 


Trigonometry. 


Courses. 

Dis- 
tances. 

Latitude. 

Departure. 

Cor. 
Lat 

Cor. 
Dep. 

Balanced.                ] 

N. 

s. 

E. 

W. 

N. 

S. 

E. 

W. 

1 

N.  45°  E. 

9.30 

6.58 

6.68 

+.01 

6.68 

6.59 

2 

S.  60° E. 

11.85 

5.92 

10.26 

+.01 

+.01 

5.93 

10.27 

3 

S.  20°  W. 

5.30 

4.98 

1.81 

-.01 

4.98 

1.80 

4 

S.  70°  W. 

10.90 

3.73 

10.24 

—.01 

3.73 

10.23 

6 

N.  310W. 

940 

8.06 

4.84 

—.01 

8.06 

4.83 

Perimeter  46.75 

14.64 

14.63 

16.84 

16.89 

14.64 

14.64 

1686 

16.86 

(156.)  When  a  field  has  been  correctly  surveyed,  and  the 
latitudes  and  departures  accurately  calculated,  the  sum  of  the 
northings  should  be  equal  to  the  sum  of  the  southings,  and  the 
sum  of  the  eastings  equal  to  the  sum  of  the  westings.  If  ihk 
northings  do  not  agree  with  the  southings,  and  the  eastings 
with  the  westings,  there  must  be  an  error  either  in  the  survey 
or  in  the  calculation.  In  the  preceding  example,  the  north- 
ings exceed  the  southings  by  one  link,  and  the  westings  ex- 
ceed the  eastings  by  five  links.  Small  errors  of  this  kind  are 
unavoidable ;  but  when  the  error  does  not  exceed  one  link  to 
a  distance  of  three  or  four  chains,  it  is  customary  to  distribute 
the  error  among  the  sides  by  the  following  proportion : 

As  the  perimeter  of  the  field, 

Is  to  the  length  of  one  of  the  sides, 

So  is  the  error  in  latitude  or  departure, 

To  the  correction  corresponding  to  that  side. 

This  correction,  when  applied  to  a  column  in  which  the  sum 
of  the  numbers  is  too  small,  is  to  be  added;  but  if  the  sum  of 
the  numbers  is  too  great,  it  is  to  be  subtracted. 

We  thus  obtain  the  corrections  in  columns  8  and  9  of  the 
preceding  table  ;  and  applying  these  corrections,  we  obtain  the 
balanced  latitudes  and  departures,  in  which  the  sums  of  the 
northings  and  southings  are  equal,  and  also  those  of  the  east- 
ings and  westings. 

As  the  computations  are  generally  carried  to  but  two  deci- 
mal places,  the  corrections  of  the  latitudes  and  departures  are 
qnly  required  to  the  nearest  link,  and  these  corrections  may 
often  be  found  by  mere  inspection  without  stating  a  formal 
proportion.  Thus,  in  the  preceding  example,  since  the  depart- 
ures require  a  correction  of  five  links,  and  the  field  has  five 
sides  which  are  not  very  unequal,  it  is  obvious  that  we  must 
make  a  correction  of  one  link  on  each  side. 


Surveying. 


109 


It  is  the  opinion  of  some  surveyors  that  when  the  error  in 
latitude  or  departure  exceeds  one  link  for  every  five  chains  of 
the  perimeter,  the  field  should  be  resurveyed ;  hut  most  sur- 
veyors do  not  attain  tp  this  degree  of  accuracy.  The  error, 
however,  should  never  exceed  one  link  to  a  distance  of  two  or 
tJiree  chains. 

(157.)  To  find  the  area  of  the  field. 

Let  ABODE  he  the  field 
to  be  measur^.  Through 
A,  the  most  western  station, 
draw  the  meridian  NS,  and 
upon  it  let  fall  the  perpen- 
diculars BF,  CG-,  DH,  EL 

Then  the  area  of  the  re- 
quired field  is  equal  to 
FBCDEI-(ABF+AEI). 

But  FBCDEI  is  equal  to 
the  sum  of  the  three  trape- 
zoids FBCG-,  OCDH,  HDEL 

Also,  if  the  sum  of  the 
parallel  sides  FB,  GrC  be  multiplied  by  FGr,  it  will  give  twice 
the  area  of  FBCG-  (Art.  87).  The  sum  of  the  sides  GC,  DH, 
multiplied  by  GrH,  gives  twice  the  area  of  GrCDH ;  and  the 
sum  of  HD,  IE,  multiplied  by  HI,  gives  twice  the  area  of 
HDEL 

Now  BF  is  the  departure  of  the  first  side,  GrC  is  the  sum  of 
the  departures  of  the  first  and  second  sides,  HD  is  the  alge- 
braic sum  oi  the  three  preceding  departures,  IE  is  the  algebraic 
sum  of  the  four  preceding  departures.  Then  the  sum  of  the 
parallel  sides  of  the  trapezoids  is  obtained  by  adding  together 
the  preceding  meridian  distances  two  by  two ;  and  if  these 
sums  are  multiplied  by  FGr,  GrH,  &c.,  which  are  the  corre- 
sponding latitudes,  it  will  give  the  double  areas  of  the  trape- 
zoids. 

(158.)  It  is  most  convenient  to  reduce  all  these  operations 
to  a  tabular  form,  according  to  the  following 

Rule. 
Having  arranged  the  balanced  latitudes  and  departures  in 


110 


Trigonometry. 


their  appropriate  columns^  draw  a  meridian  through  the  most 
eastern  or  western  station  of  the  survey,  and,  calling-  this  the 
first  station,  form  a  column  of  double  meridian  distances. 

The  double  meridian  distance  of  the  first  side  is  equal  to 
its  departure  ;  and  the  double  meridian  distance  of  any  side 
is  equal  to  the  double  meridian  distance  of  the  preceding  side^ 
plus  its  departure,  plus  the  departure  of  the  side  itself 

Multiply  each  double  meridian  distance  by  its  correspond- 
ing northing  or  southing,  and  place  the  prodm^  in  the  column 
of  north  or  south  areas.  The  difference  betmeen  the  sum  of 
the  north  areas  and  the  sum  of  the  south  areas  will  be  double 
the  area  of  the  field. 

It  must  be  borne  in  mind  that  by  the  term  plus  in  this  rule 
is  to  be  understood  the  algebraic  sum.  Hence,  when  the 
double  meridian  distance  and  the  departure  are  both  east  or 
both  west,  they  must  be  added  together  ;  but  if  one  be  east 
and  the  other  west,  the  one  must  be  subtracted  from  the  other. 

The  double  meridian  distance  of  the  last  side  should  always 
be  equal  to  the  departure  for  that  side.  This  coincidence  af- 
fords a  check  against  any  mistake  in  forming  the  column  of 
double  meridian  distances. 

The  preceding  example  will  then  be  completed  as  follows  : 


N. 

s. 

E. 

w. 

D.M.D.  1   N.  Areas. 

S.  Areas. 

1 

6.58 

6.59 

6.59 

43.3622 

2 

5.93 

10.27 

23.45 

139.0585 

8 

4.98 

1.80 

31.92 

158.9616 

4 

3.73 

10.23 

19.89 

74.1897 

5 

8.06 

4.83 

4.83 

38.9298 

82.2920 

372.2098 

Twice  the  figure  FBCDEI  is  372.2098  square  chains. 
Twice  the  figure  FBAEI     is    82.2920  " 

The  difference  is     ...     .  289.9178  " 

Therefore  the  area  of  the  field  is  144.9589  square  chains,  or 

14.49589  acres,  which  is  equal  to  14  acres,  1  rood,  39  perches. 
Ex.  2.  It  is  required  to  find  the  contents  of  a  tract  of  land 

of  which  the  following  are  the  field  notes : 


SURVEVINO. 


Ill 


Sta- 
tions. 

Bearings. 

Distances, 

1 

N.  50°  30'  E. 

16.50  chains. 

2 

S.  68°  15'  E. 

14.20      " 

3 

S.     9°  45'  E. 

8.45       '^ 

4 

S.  21°    0'  W. 

6.84      " 

5 

S.  73°  30'  W. 

12.31      " 

6 

N.  78°  15'  W. 

9.76      " 

7 

N.  17°    0'  W. 

11.64      " 

Calculation. 


Courses. 

Dist 

Dif.  Lat 

De|>nrture. 

1               Balanced. 

DM. 

S.  Areas. 

1 

N. 

s. 

K.        W. 

Co,.|    j,_ 

s. 

K. 

w. 

D. 

N.  5()0  3(7  E. 

16.50,10.50 

12.73 

.03  10.47 

12.70 

1270 

132.9690 

2 

S.  680  15'  E. 

14.20 

5.26 

13.19 

.03 

5.29 

13.16 

38.56 

203.9824 

3 

S.    90  45'  E. 

8.45 

8.33 

1.43 

.01 

8.34 

1.42 

.53.14 

443.1876 

4 

S.  210    O'VV. 

6.84 

6.39 

2.45 

.01 

6.40 

2.46 

52.10 

333.4400 

6 

S.  730  30'  w. 

12.31 

3.50 

11.80 

,02 

3,52 

11.82  37.82 

133.1264 

6 

N.  780  15'  W. 

9.76 

1.99 

9.56 

.02 

1.97 

9..'>8 

16.42'  32.3474 

7 

N.  170    CW. 

11.64111. 13 

3.40 

.02 

11.11 

3.42 

3.42  37.9962 

79.70,23.6223.48 

27.35 

27.21 

23  5o 

23.55 

27.28 

27.28 

,20  !.:<  26 

11137364 

1  Error  .14 

Error  .14 

203.3126 

Ans.,  45  A.,  2  R.,  3  P. 
Ex.  3.  Required  the  area  of  a 
tract  of  land  of  which  the  follow- 
ing are  the  field  notes  : 


2)910.4238 
,     455.2119. 


Sta- 
tions. 

Bearings. 

Distances. 

1 

N.  58°  45'  E. 

19.84  chains. 

2 

N.  39°  30'  E. 

10.45 

3 

S.  45°  15'  E.  i37.26 

4 

S.  52°  30'  W.'21.53 

5 

S.  34°    0'  E. 

9.12 

6 

S.  m^"  15'  W. 

27.69 

7 

N.  12°  45'  E. 

24.31 

8 

N.  48°  15'  W.  24.60 

Ans.,  130  A.,  2  R.,  23  P. 
Ex.  4.  Required  the  area  of  a  piece  of  land  from  the  follow- 


ing  field  notes 


112 


Trigonometry. 


stations. 

Bearings. 

Distances. 

1 

N.    5°  15'  E. 

15.17  chains. 

2 

N.  45°  45'  E. 

16.83      " 

3 

N.  32°    0'  W. 

14.26      " 

4 

N.  88°  30'  E. 

19.54      " 

5 

S.  28°  15'  E. 

17.92      " 

6 

S.  40°  45'  W. 

9.71      " 

7 

S,  31°  30'  E. 

22.65      " 

8 

S.  14°    0'  W. 

18.39      " 

9 

S.  82°  45'  W. 

24.80      " 

10 

N.  23°  15'  W. 

26.31      " 

Ans.,  173  A.,  0  R.,  23  P. 
Ex.  5.  Required  the  area  of  a  field  from  the  following  notes 


Stations. 


1 

2 
3 
4 
5 

6 

7 

8 

9 

10 

11 

12 


Bearings. 


Distances. 


N.  32°  15'  E. 
N.  17°  45'  E. 
S.  81°  30'  E. 
S.  9°  45'  W. 
S.  43°  0'  E. 
N.  25°  30' 
S.  78°  15' 
S.  5°  45' 
S.  37°  30' 
N.  69°  0' 
S.  74°  15' 
N.  27°  30' 


E. 

E. 

W. 

W. 

W. 

W. 

W. 


28.74  chains. 

21.59  " 

13.38  " 

11.92  '' 

19.65  " 

17.26  " 

18.87  " 

31.41  " 

26.13  '« 

23.86  " 

20.^1  " 

23.20  " 


A71S.,  304  A.,  2  E,.,  9  P. 
Ex.  6.  Required  the  area  of  a  field  from  the  following  notes; 


Stations. 

Bearings. 

Distances. 

1 

N.  36°  15'  E. 

24.73  chains. 

2 

N.    7°  45'  E. 

11.58      ^' 

3 

N.  79°  30'  E. 

15.39      " 

4 

S.  86°  45'  E. 

20.56      " 

5 

S.  12°  15'  W. 

18.14      " 

6 

S.  25°    0'  E. 

21.92      " 

7 

S.  58°  30'  W. 

29.27      " 

8 

N.  34°    0'  W. 

19.81      '' 

9 

N.  81°  15'  W. 

21.24      " 

ilw.s.,  179  A.,  1  R.,  6  P. 


Surveying. 


113 


(159 )  The  field  notes  from  which  the  area  is  to  be  com- 
puted may  be  imperfect.  There  may  be  obstacles  which  pre- 
vent the  measuring  of  one  side,  or  the  notes  may  be  defaced 
80  as  to  render  some  of  the  numbers  illegible.  If  the  bearings 
and  lengths  of  all  the  sides  of  a  field  except  one  are  given,  the 
remaining  side  may  easily  be  found  by  calculation.  For  the 
diiference  between  the  sum  of  the  northings  and  the  sum  of 
the  southings  of  the  given  sides  will  be  the  northing  or  south- 
ing of  the  remaining  side  ;  and  the  difference  between  the  sum 
of  the  eastings  and  the  sum  of  the  westings  of  the  given  sides 
will  be  the  easting  or  westing  of  the  remaining  side.  Having, 
then,  the  difference  of  latitude  and  departure  of  the  required 
side,  its  length  and  direction  are  easily  found  by  Trigonome- 
try (Art.  47). 

Ex.  Griven  the  bearings  and  lengths  of  the  sides  of  a  tract 
of  land  as  follows  : 


Stations. 

Bearings. 

Distances. 

1 

N.  18°  15'  E. 

8.93  chains. 

2 

N.  79°  45'  E. 

15.64       " 

3 

S.  25°    0'  E. 

14.27       " 

4 

Unknown. 

Unknown. 

5 

N.  87°  30'  W. 

18.52  chains. 

6 

N.  41°  15'  W. 

12.18       " 

Kequired  the  bearing  and  distance  of  the  fourth  side. 

Ans.,  S.  15°  33'  E.,  distance  8.62  chains.    • 

(160.)  There  is  another  method  of  finding  the  area  of  a  field 
which  may  be  practiced  when  great  accuracy  is  not  required 
It  consists  in  first  drawing  a  plan  of  the  field,  as  in  Art.  153 
then  dividing  the  field  into 
triangles  by  diagonal  lines, 
and  measuring  the  bases  and 
perpendiculars  of  the  triangles 
upon  the  same  scale  of  equal 
parts  by  which  the  plot  was 
drawn.  Thas,  if  we  take  Ex. 
1,  and  draw  the  diagonals  AC, 
AD,  the  field  will  be  divided 
into  three  triangles,  whose  area 
s  easily  found  when  we  know 

H 


114 


Trigonometry. 


the  diagonals  AC,  AD,  and  the  perpendiculars  BF,  DGr,  EH 
The  diagonal  AC  is  found  by  measurement  upon  the  scale  of 
equal  parts  to  be  16.87 ;  the  diagonal  AD  is  15.67  ;  the  perpen- 
licular  BF  is  6.30 ;  DO  is  4.92  ;  and  EH  is  6.42.     Hence 

the  triangle  ABC=16.87X3.15=  53.14 
"  "  ADC=16.87X2.46=  41.50 
"         ''      ADE=15.67x3.21=  50.30 


the  figure  ABCDE  =144.94  sq.  chains. 

This  method  of  finding  the  area  of  a  field  is  very  expedi- 
tious, and  when  the  plot  is  carefully  drawn,  may  afford  results 
sufficiently  precise  for  many  purposes. 

(161.)  To  survey  an  irregular  boundary  by  means  of  off- 
sets. 

When  the  boundaries  of  a  field  are  very  irregular,  like  a 
river  or  lake  shore,  it  is  generally  best  to  run  a  straight  line, 
coming  as  near  as  is  convenient  to  the  true  boundary,  and 
measure  the  perpendicular  distances  of  the  prominent  points 
of  the  boundary  from  this  line. 

Let  ABCD  be  a  piece  of  land  to  be  surveyed  ^  the  land  be- 
ing bounded  on  the  east  by  a  lake,  and  on  the  west  by  a  creek 
We  select  stations  A,  B,  C, 
D,  so  as  to  form  a  polygon 
which  shall  embrace  most  of 
^he  proposed  field,  and  find 
its  area.  We  then  measure 
perpendiculars  aa'^  bb\  cc', 
&c.,  as  also  the  distances  Aa, 
ab,  be,  &c.  Then,  consider- 
ing the  spaces  Aaa\  abb'a', 
&c.,  as  triangles  or  trapezoids, 
their  area  may  be  computed ; 
and,  adding  these  areas  to  the 
figure  ABCD,  we  shall  obtain 
the  area  of  the  proposed  field  nearly. 

(162.)  To  determine  the  bearing  and  distance  from  onf 
point  to  another  by  means  of  a  series  of  triangles. 

When  it  is  required  to  find  the  distance  between  two  points 
remote  from  each  oth^.r,  we  form  a  series  of  triangles  such  that 


Surveying. 


llfl 


the  first  and  second  triangles  may  have  one  side  in  common  ; 
the  second  and  thii*d,  also,  one  side  in  common  ;  the  third  and 
fourth,  &c.  We  then  measure  one  side  of  the  first  triangle 
for  a  base  line,  and  all  the  angles  in  each  of  the  triangles. 
These  data  ard  sufficient  to  determine  the  length  of  the  sides 
of  each  triangle  ;  for  in  the  first  triangle  we  have  one  side  and 
the  angles  to  find  the  other  sides.  When  these  are  found,  we 
shall  have  one  side  and  all  the  angles  of  a  second  triangle  to 
find  the  other  sides.  In  the  same  manner,  we  may  calculate 
the  dimensions  of  the  third  triangle,  the  fourth,  and  so  on.  We 
shall  illustrate  this  method  hy  an  example  taken  from  the 
Coast  Survey  of  the  United  States. 

The  object  here  is  to  make  a  survey  of  Chesapeake  Bay  and 
its  vicinity ;  to  determine  with  the 
utmost  precision  the  position  of  the 
most  prominent  points  of  the  country, 
to  which  subordinate  points  may  be 
referred,  and  thus  a  perfect  map  of 
the  country  be  obtained.  According- 
ly, a  level  spot  of  ground  was  select- 
ed on  the  eastern  side  of  the  bay,  on 
Kent  Island,  where  a  base  line,  AB, 
of  more  than  five  miles  in  length, 
was  measured  with  every  precaution. 
A  station,  C,  was  also  selected  upon 
the  other  side  of  the  bay,  near  An- 
napolis, so  situated  that  it  was  visi- 
ble from  A  and  B.  The  three  angles 
of  the  triangle  ABC  were  then  meas- 
ured with  a  large  theodolite,  after 
which  the  length  of  BC  may  be  com- 
puted. A  fourth  station,  D,  is  now  taken  on  the  western  shore 
of  the  bay,  visible  from  C  and  B,  and  all  the  angles  of  the  tri- 
angle BCD  are  measured,  when  the  line  BD  can  be  computed. 
A  fifth  station,  E,  is  now  taken  on  an  island  near  the  eastern 
shore,  visible  both  from  B  and  D,  and  all  the  angles  of  the  tri- 
angle BDE  are  measured,  when  DE  can  be  computed.  Also, 
all  the  angles  of  the  triangle  DEF  are  measured,  and  EF  is 
computed.     Then  all  the  angles  of  the  triangle  EFGr  are  mea«- 


116  Trigonometry. 

ured,  and  FGr  is  computed.  So,  also,  all  the  angles  of  the  tri- 
angle FGH  are  measured,  and  G-H  is  computed;  and  thus  a 
chain  of  triangles  may  be  extended  along  the  entire  coast  of 
the  United  States.  To  test  the  accuracy  of  the  work,  it  is 
common  to  measure  a  side  in  one  of  the  triangles  remote  from 
the  first  base,  and  compare  its  measured  length  with  that  de- 
duced by  computation  from  the  entire  series  of  triangles.  This 
line  is  called  a  base  of  verification.  Such  a  base  has  been 
measured  on  Long  Island  ;  and,  indeed,  several  bases  have 
been  measured  on  different  points  of  the  coast.  These  are  all 
connected  by  a  triangulation,  and  thus  the  length  of  a  side  in 
any  triangle  may  be  deduced  from  more  than  one  base  line, 
and  the  agreement  of  these  results  is  a  test  of  the  accuracy  of 
the  entire  work.  Thus  the  length  of  one  of  the  sides  of  a  tri- 
angle which  was  twelve  miles,  as  deduced  from  the  Kent  Island 
base,  differed  only  twenty  inches  from  that  derived  from  the 
Long  Island  base,  distant  two  hundred  miles. 

The  superiority  of  this  method  of  surveying  arises  from  the 
circumstance  that  it  is  necessary  to  measure  but  a  small  num- 
ber of  base  lines  along  a  coast  of  a  thousand  or  more  miles  in 
extent ;  and  for  these  the  most  favorable  ground  may  be  se- 
lected any  where  in  the  vicinity  of  the  system  of  triangles. 
All  the  other  quantities  measured  are  angles ;  and  the  pre- 
cision "of  these  measurements  is  not  at  all  impaired  by  the  in- 
equalities of  the  surface  of  the  ground.  Indeed,  mountainous 
countries  afford  peculiar  facilities  for  a  trigonometrical  survey, 
since  they  present  heights  of  ground  visible  to  a  great  distance, 
and  thus  permit  the  formation  of  triangles  of  very  large  di- 
mensions. 

(163.)  To  divide  an  irregular  piece  of  land  into  any  two 
given  parts. 

We  first  run  a  line,  by  estimation,  as  near  as  may  be  to  th. 
required  division  line,  and  compute  the  area  thus  cut  off.  It 
this  is  found  too  large  or  too  small,  we  add  or  subtract  a  tri- 
angle, or  some  other  figure,  as  the  case  may  require.  Sup- 
pose it  is  required  to  divide  the  field  ABODE  FGr  HI  into  two 
equal  parts,  by  a  line  IL,  running  from  the  corner  I  to  the 
opposite  side  CD.  We  first  draw  a  line  from  I  to  D,  and  com- 
pute the  area  of  the  part  DEFGrHI ;  and,  knowing  the  area 


Surveying.  117 

of  the  entire  field,  we  learn  the  area  which  must  be  containal 
in  the  triangle  DIL,  in  order  that  ^ 

IL  may  divide  the  field  into  two  /"^^-^C 

equal  parts.    Having  the  bearings  /  \ 

and  distances  of  the  sides  DEjEF,  /  \  » 

&c.,  we  can  compute  the  bearing         /  »  XlC 

and  distance  of  DI.    Thus  the  an-  ^/  --"''l!!^--'^ 

gle  IDK  is  known;  and,  having       ^^"^^^.^^i^^^^^^^'''^^^      -\d 
the  hypothenuse  ID,  we  can  com-  I|  >/ 

pute  the  length  of  the  perpendicu-  /^ 

lar  IK  let  fall  on  CD.     Now  the  I  \ 

base  of  a  triangle  must  be  equal  -^y  \ 

to  its  area  divided  by  half  the  al-  \ \ 

titude.     Hence,  if  we  divide  the  ^  ^ 

area  of  the  triangle  DIL  by  half  of  IK,  it  will  give  DL. 

In  a  similar  manner  we  might  proceed  if  it  was  required  to 
4ivide  a  tract  of  land  into  any  two  given  parts. 

Variation  of  the  Needle. 

(164.)  The  line  indicated  by  a  magnetic  needle,  when  free- 
ly supported  and  allowed  to  come  to  a  state  of  rest,  is  called 
the  magnetic  meridian.  This  does  not  generally  coincide  with 
the  astronomical  meridian,  which  is  a  true  north  and  south 
line. 

The  angle  which  the  magnetic  meridian  makes  with  the 
true  meridian  is  called  the  variation  of  the  needle,  and  is  said 
to  be  east  or  west,  according  as  the  north  end  of  the  needle 
points  east  or  west  of  the  north  pole  of  the  earth. 

The  variation  of  the  needle  is  different  in  different  parts  of 
the  earth.  In  some  parts  of  the  United  States  it  is  10°  we^^t, 
and  in  others  10°  east,  while  at  other  places  the  variation  has 
every  intermediate  value.  Even  at  the  same  place,  the  varia- 
tion does  not  remain  constant  for  any  length  of  time.  Hence 
it  is  necessary  frequently  to  determine  the  amount  of  the  varia- 
tion, which  is  easily  done  when  we  know  the  position  of  the 
true  meridian.  The  latter  can  only  be  determined  by  astro- 
nomical observations.  The  best  method  is  by  observations  of 
the  pole  star.  If  this  star  were  exactly  at  the  pole,  it  would 
alw  ays  be  on  the  meridian ;  but,  being  at  a  distance  of  about 


118 


Trigonometry. 


a  degree  and  a  half  from  the  pole,  it  revolves  about  the  polo  in 
a  small  circle  in  a  little  less  than  24  hours.  In  about  six  hours 
from  its  passing  the  meridian  above  the  pole,  it  attains  its 
greatest  distance  west  of  the  meridian  ;  in  about  six  hours 
more  it  is  on  the  meridian  beneath  the  pole  ;  and  in  about  six 
hours  more  it  attains  its  greatest  distance  «ast  of  the  meridian. 
If  the  star  can  be  observed  at  the  instant  when  it  is  on  the 
meridian,  either  above  or  below  the  pole,  a  true  north  and  south 
line  may  be  obtained. 

(165.)  The  following  table  ^hows  the  time  of  the  pole  star's 
passing  the  meridian  above  the  pole  for  every  fifth  day  of  the 
year : 


l8t  Day. 

6tli  Day. 

lull  Day. 

inth  Day. 

'21st  Day. 

2(;tli  Dny. 

h.     m. 

h.     ni. 

h.    m. 

h     m 

li.      m. 

h.     111. 

January    .  .  . 

6  20  P.M. 

6     0  P.M. 

5  41   P.M. 

5  2l'  P.M. 

5    1  P.M. 

4  42  P.M. 

February.  .  . 

4  18     « 

3  58     " 

3  39     " 

3  19     " 

3    0" 

2  40     " 

March    .... 

2  28     " 

2    8" 

1  49     '« 

1  29     " 

1    9     " 

0  50     " 

April 

0  2()     " 

0    7     " 

11  47  A.M. 

11  27  A.M. 

11    8  A.M. 

10  48  A.M. 

M»y 

]0  28  A.M. 

10    9  A.M. 

9  49     " 

9  29     '• 

9    9     " 

8  50     " 

June 

8  26     " 

8    7" 

7  47     " 

7  27     " 

7    8" 

6  48     " 

July 

6  28     " 

6    9" 

5  49     " 

5  29     " 

5  10     " 

4  50     " 

August  .... 

4  27     " 

4    7     •» 

3  47      " 

3  27     " 

3    8" 

2  48     " 

September .  . 

2  25     " 

2    5     " 

1  45     " 

1  26     " 

1     6     " 

0  46      " 

October    .  .  . 

0  26     " 

0    7" 

11  43  P.M. 

11  24  P.M. 

11     4  P.M. 

10  44  P.M. 

November  .  . 

10  21  P.M. 

10    1  P.M. 

9  41      " 

9  22     " 

9    2     '• 

8  42     " 

December  .  . 

8  23     " 

8    3" 

7  43      " 

7  24     •' 

7    4" 

6  44      " 

If  the  pole  star  passes  the  meridian  in  the  daytime,  it  can 
not  be  observed  without  a  good  telescope  ;  but  11'''  58'"  after 
the  dates  in  the  above  table,  the  star  will  be  on  the  meridian 
below  the  pole,  and  during  the  whole  year,  except  in  summer, 
the  pole  star  may  be  seen  with  the  naked  eye  on  the  merid- 
ian either  above  or  below  the  pole.  These  observations  are 
best  made  with  a  theodolite,  but  they  may  be  made  with  a 
common  compass.  At  5''"  SO"""  after  the  dates  in  the  above 
table,  the  star  will  have  attained  its  greatest  distance  west  of 
the  meridian  ;  and  5""  59'"'  before  these  dates,  it  will  be  at  its 
greatest  distance  east  of  the  meridian.  In  summer,  therefore, 
we  may  observe  the  greatest  eastern  elongation  of  the  pole  star, 
at  which  time  the  star  is  1°  6^'  east  of  the  true  meridian  for 
all  places  in  the  neighborhood  of  New  York.  Making  this  al- 
lowance, a  true  meridian  is  easily  obtained ;  after  which,  the 
variation  of  the  needle  is  determined  by  placing  a  compass 
upon  this  line,  turning  the  sights  in  the  same  direction,  and 
noting  the  angle  shown  by  the  needle. 

The  following  table  shows  the  angle  which  the  plane  of  the 


Surveying. 


119 


meridian  makes  with  a  vertical  plane  passing  through  the  pole 
star,  when  at  its  greatest  eastern  or  western  elongation,  for 
any  latitude  from  30^  to  44°. 


Lat.  30^ 

Lat.  32° 

Lat.  34° 

Lat.  36° 

Lat.  380 

1°  41' 

1M3' 

r45' 

IMS' 

V  51' 

Lat.  40°  I  Lat.  42°  |Lat.  44'» 
1°  54'  I  1°  58'     2°  2' 


(166.)  The  variation  of  the  needle,  in  1840,  for  several  parts 
^f  the  United  States,  was  as  follows  : 


Burlington,  Yt.  . 
Boston,  Mass.  . 
Albany,  N.  Y.  . 
New  Haven,  Ct. 
New  York  City 
Philadelphia    .  . 


9°  27' 
9°  12' 
6°  58' 
6°  13' 
5°  34' 
40    s' 


w. 
w. 
w. 
w. 
w. 
w. 


Washington  City  .  1°  20'  W. 


Buffalo,  N.  Y.  . 
Cleveland,  Ohio 
Detroit,  Mich.  . 
Charleston,  S.  C. 
Cincinnati,  Ohio 
Mobile,  Ala.  .  . 
St.  Louis,  Mo.    . 


1°  37' 
0°  19' 
1°56' 

2°  44' 
4°  46' 

7°  5' 
8°  37' 


W. 

E. 
E. 
E. 
E. 
E. 

e: 


Since  1840,  the  variation  in  New  England  has  increased 
about  five  minutes  annually  ;  in  New  York  and  Pennsylvania 
it  has  increased  from  three  to  four  minutes  annually.  In  the 
Western  States  it  decreases  at  about  the  same  rate,  and  in  the 
Southern  States  it. decreases  about  two  minutes  annually. 


LEVELING. 

(167.)  Leveling'  is  the  art  of  determining  the  difference  ol 
icvel  between  two  or  more  places. 

The  surface  of  an  expanse  of  tranquil  water,  or  any  surface 
parallel  to  it,  is  called  a  level  surface.  Points  situated  in  a 
level  surface  are  said  to  be  on  the  same  levels  and  a  line  traced 
on  such  a  surface  is  called  a  line  of  true  level. 

On  account  of  the  globular  figure  of  the  earth,  a  level  sur- 
face is  not  a  plane  surface.  It  is  nearly  spherical ;  and  in  the 
common  operations  of  leveling  it  is  regarded  as  perfectly  so. 
Hence  every  point  of  a  level  surface  is  regarded  as  at  the  same 
distance  from  the  center  of  the  earth ;  and  the  difference  of 
level  of  two  places  is  the  difference  between  their  distances 
from  the  center. 

A  line  of  apparent  level  is  a  straight  line  tangent  to  the  sur- 
face of  the  earth. 

Thus,  if  AB  represent  the  surface  of  the  ocean,  the  two 
places  A  and  B  are  said  to  be  on  the  same  level ;  but  if  AP 


i* 


120 


Trigonometry 


be  drawn  tangent  to  the  arc  AB  at  A,  then  AD  is  a  line  oi 
apparent  level.  j^ 

This  is  the  line  which  is  indicated  by  a 
leveling  instrument  placed  at  A.  The  theod- 
olite may  he  employed  for  tracing  horizontal 
lines ;  hut  if  nothing  further  were  required, 
there  would  he  no  occasion  for  graduated  cir- 
cles, and  several  parts  of  the  theodolite  might 
be  dispensed  with.  A  leveling  instrument,  therefore,  usually 
consists  of  a  large  spirit  level  attached  to  a  telescope,  mounted 
upon  a  stand  in  a  manner  similar  to  the  theodolite. 

(168.)  The  surveyor  should  also  be  provided  with  a  pair  of 
leveling-  staves.  A  leveling  staff  consists  of  a 
rectangular  bar  of  wood  six  feet  in  length,  di-  A|  L^^B 
vided  to  inches  and  sometimes  tenths  of  an 
inch,  and  having  a  groove  running  its  entire 
length.  A  smaller  staff  of  the  same  length, 
called  a  slide,  also  divided  into  inches,  is  in- 
serted in  this  groove,  and  moves  freely  along  it. 

At  the  upper  end  of  the  slide  is  a  rectangu- 
lar board  called  a  vane,  AB,  about  six  inches 
wide.  The  vane  is  divided  into  four  equal  parts  by  two  lines, 
one  horizontal  and  the  other  vertical.  Two  opposite  parts  of 
the  vane  are  painted  white,  and  the  other  two  black,  in  order 
that  they  may  be  distinguished  at  a  great  distance. 

To  find  the  difference  of  level  between  any  two  points. 
(169.)  Set  up  the  leveling  staves  perpendicular  to  the  hori- 
zon, and  at  equal  distances  from  the  leveling  instrument. 
Having  adjusted  the  level  by  means  of  the  proper  screws,  turn 
the  telescope  to  one  of  the  staves,  and  direct  an  assistant  to 
slide  up  the  vane  until  the  line  AB  coincides  with  the  center 
of  the  telescope,  and  note  the  height  of  this  line  from  the 
ground.  Turn  the  telescope  to  the  other  staff,  and  repeat  the 
same  operation.  Level  in  the  same  manner  from  the  second 
station  to  the  third,  from  the  third  to  the  fourth,  &c.  Then 
the  difference  between  the  sum  of  the  heights  at  the  back  sta- 
tions and  at  the  forward  stations  will  be  equal  to  the  difference 
of  level  between  the  first  station  and  the  last. 


N 


—  E 


Purveying.  ,  121 

If  we  wish  to  level  fiom  A  to  E,  we  set  up  the  staves  at  a 
convenient  distance, 

AC,  and  midway  be-  I ^ •  H 

tween  them  place  the    ^^^^^  ^  ^|" "T 

level    B.       Observe       '"^^^^^^i^-^^^^^^^^^^^M 
where  the  line  of  lev-  ~  ^^^^^ 

el,  FG,  cuts  the  rods,  and  note  the  heights  AF,  CGr.  Their 
difference  is  the  diiference  of  level  between  the  first  and  second 
stations.  Take  up  the  level  and  place  it  at  D,  midway  be- 
tween the  rods  C  and  E,  and  observe  where  the  line  of  level, 
[TI,  cuts  the  rods,  and  note  the  heights  CH,  El. 

Then  FA— CGr=:the  ascent  from  A  to  C, 

and  CH— EI  =the  ascent  from  C  to  E. 

Therefore  (FA+CH)  — (Ca+EI)=the  entire  ascent  from 
A  to  E  ;  and  in  the  same  manner  we  may  find  the  difference 
of  level  for  any  distance ;  that  is,  the  difference  between  the 
sum  of  the  heights  at  the  back  stations  and  at  the  forward 
stations  is  equal  to  the  difference  of  level  between  the  first  sta- 
tion and  the  last, 

(170.)  The  following  is  a  copy  of  the  field  notes  for  running 
a  level  from  A  to  E  : 


%ck  sights. 

Feet.    Inches. 

0      4 

Fore 

Feet. 

3 

sights. 

Inches. 

2 

5     10 

5 

7 

4      2 

4 

3 

5       6 

1 

2 

4     11 

3 

2 

4      7 

1 

3 

6       1 

2 

0 

Sum  31       5  .am  20       7 

The  back  sights  being  greater  in  amount  than  the  forward 
sights,  it  is  evident  that  E  is  higher  than  A  by  10  feet  10 
inches. 

The  heights  indicated  by  the  leveling  staves  are  sometimes 
read  off  by  the  assistant,  but  it  is  better  for  the  observer  to 
read  off  the  quantities  himself  through  the  telescope  of  his 
leveling  instrument.  This  may  easily  be  done  provided  the 
graduation  of  the  staff  is  perfectly  listinct ;  and  in  that  case  it 


122 


Trigonometry. 


is  only  necessary  to  rely  upon  the  assistant  to  hold  the  staff 
perpendicularly.  To  enable  him  to  do  this,  a  small  plummet  is 
suspended  in  a  groove  cut  in  the  side  of  the  staff. 

(171.)   It  must  be  observed  that  the  lines  GrF,  HI  are  lines 
of  apparent  level,  and 


-^_    ^i T 


not  oitrue  level ;  nev- 
ertheless, we  shall  ob- 
tain the  true  differ- 
ence of  level  between 


A.  and  E  by  this  method  if  the  leveling  instrument  is  placed 
midway  between  the  leveling  staves,  because  the  points  (ir  and 
F  will  in  that  case  be  at  equal  distances  from  the  earth's  cen^ 
ter.  If  the  level  is  not  placed  midway  between  the  staves, 
then  we  must  apply  a  correction  for  the  difference  between 
the  true  and  apparent  level. 

(172.)  To  find  the  difference  between  the  true  and  apparent 
level. 

Let  C  be  the  center  of  the  earth,  AB  a  portion  of  its  surface, 
and  AD  a  tangent  to  the  earth's  surface  at 
A;   then  BD  is  the  difference  between  the    — :- 
true  and  apparent  level  for  the  distance  AD. 

Now,  by  Geom.,  Prop.  11,  B.  IV., 


CD^=AC^-}-AD^ 


Hence 
and 


CD  =  VAC^+AD^ 
BD=VAC'-fAD^-BC. 


If  we  put  R  =  BC,  the  radius  of  the  earth, 
^=AD, 
and  A=BD,  the  difference  between  the  true  and  ap- 

parent level,  we  shall  have 


that  is,  to  find  the  difference  between  the  true  and  apparent 
level  for  any  distance,  add  the  square  of  the  distance  to  the 
square  of  the  earth'' s  radius^  extract  the  square  root  of  the 
sum,  and  subtract  the  radius  of  the  earth. 

If  BD  represent  a  mountain,  or  other  elevated  object,  then 
AD  will  represent  the  distance  at  which  it  can  be  seen  in  con 
sequence  of  the  curvature  of  the  earth. 


whence  h=-^^;^\ 


Surveying.  123 

Ex.  1.  If  the  diameter  of  the  earth  be  7912  miles,  and  if 
Mount  x-Etna  can  be  seen  at  sea  126  miles,  what  is  its  height? 

Ans.,  2  miles. 

Ex.  2.  If  a  straight  line  from  the  summit  o£  Chimborazo 
touch  the  surface  of  the  ocean  at  the  distance  of  179  miles, 
what  is  the  height  of  the  mountain  ?  Ans.,  4.05  miles. 

From  the  preceding  formula  we  obtain 

-R^+2IU+/i^ 
that  is,  6f =2R/i+/i'. 

But  in  the  common  operations  of  leveling,  h  is  very  small  m 
comparison  with  the  radius  of  the  earth,  and  A^  is  very  small 
in  comparison  with  2R/i.     If  we  neglect  the  term  A"*,  we  have 

2R 

that  is,  the  difference  between  the  true  and  apparent  level  is 
nearly  equal  to  the  square  of  the  distance  divided  by  the  di^ 
ameter  of  the  earth. 

Ex.  1.  What  is  the  difference  between  the  true  and  apparent 
level  for  one  mile,  supposing  the  diameter  of  the  earth  to  be 
7912  miles?  Ans.,  8.008  inches,  or  8  inches  nearly. 

Ex.  2.  What  is  the  difference  between  the  true  and  apparent 
level  for  half  a  mile  ?  Ans.,  2  inches. 

In  the  equation  h=-^,  since  2R  is  a  constant  quantity,  h 

varies  as  d"" ;  that  is,  the  difference  between  the  true  and  ap- 
parent  level  varies  as  the  square  of  the  distance. 

Hence,  the  difference  for  1  mile  being  8  inches.     Ft.  in 
the  difference  for  2  miles  is  8x2'=  32  inches=  2  8. 
"  "         3        ''       8x3'=  72     "      =  6  0. 

*'  "         4        "       8x4^=128     ''      =10  8. 

*'  "        5        "       8x5'=200     "     =16  8. 

"  "         6        ^'       8x6'=288     "      =24  0,  &c. 

Topog'rap^iical  Maps. 
(173.)  It  is  sometimes  required  to  determine  and  represent 
upon  a  map  the  undulations  and  inequalities  in  the  surface  of 


124 


Trigonometry. 


a  tract  of  land.  Such  a  map  sliould  give  a  complete  view  ol 
the  ground,  so  as  to  afford  the  means  for  an  appropriate  loca- 
tion of  buildings  or  extensive  works.  For  this  purpose,  we 
suppose  the  surface  of  the  ground  to  be  intersected  by  a  num- 
ber of  horizontal  planes,  at  equal  distances  from  each  other. 
The  lines  in  which  these  planes  meet  the  surface  of  the  ground, 
being  transferred  to  paper,  will  indicate  the  variations  in  the 
inclination  of  the  ground  ;  for  it  is  obvious  that  the  curves  will 
be  nearer  together  or  further  apart,  according  as  the  ascent  is 
steep  or  gentle. 

Thus,  let  ABCD  be  a  tract  of  broken  ground,  divided  by  a 
stream,  EF,  the  ascent  being  rapid  on  each  bank,  the  ground 
swellinof  to  a  hill  A.  E  r 


at  G,  and  also  at 
H.  It  is  required 
to  represent  these 
inequalities  upon 


paper, 


so    as    to 


give  an  exact  idea 
of  the  face  of  the 
ground.  The  low- 
est point  of  the 
ground  is  at  F. 
Suppose  the  tract 

to  be  intersected  C  y  '     ~~  D 

by  a  horizontal  plane  four  feet  above  F,  and  let  this  plane  in- 
tersect the  surface  of  the  ground  in  the  undulating  lines  marked 
4,  one  on  each  side  of  the  stream.  Suppose  a  second  horizon- 
tal plane  to  be  drawn  eight  feet  above  F,  and  let  it  intersect 
the  surface  of  the  ground  in  the  lines  marked  8.  Let  other 
horizontal  planes  be  drawn  at  a  distance  of  12,  16,  20,  24, 
&c.,  feet  above  the  point  F.  The  projection  of  these  lines  of 
level  upon  paper  shows  at  a  glance  the  outline  of  the  tract. 
We  perceive  that  on  the  right  bank  of  the  stream  the  ground 
rises  more  rapidly  on  the  upper  than  on  the  lower  portion  of 
the  map,  as  is  shown  by  the  lines  of  level  being  nearer  to  one 
another.  On  the  right  bank  of  the  stream  the  ascent  is  unin- 
terrupted until  we  reach  Gr,  which  is  the  summit  of  the  hill. 
Beyond  Gr  the  ground  descends  again  toward  B.     On  the  left 


125 

Dank  of  the  stream  the  ground  rises  to  H ;  hut  toward  A  the 
level  line  of  12  feet  divides  into  two  branches,  and  between 
them  the  ground  is  nearly  level. 

(174.)  The  surveys  requisite  for  the  construction  of  such  a 
map  may  be  made  with  a  theodolite  or  common  level. 

The  object  is  to  trace  a  series  of  level  lines  upon  the  surface 
of  the  ground.  For  this  purpose  we  may  select  any  point  on 
the  surface  of  a  hill,  place  the  level  there,  and  run  a  level  line 
around  the  hill,  measuring  the  distances,  and  also  the  angles, 
at  every  change  of  direction.  We  may  then  select  a  second 
point  at  any  convenient  distance  above  or  below  the  former, 
and  trace  a  second  level  line  around  the  hill,  and  so  on  for  as 
many  curves  as  may  be  thought  necessary.  Such  a  method, 
however,  would  not  always  be  most  convenient  in  practice. 

(175.)  The  following  method  may  sometimes  be  preferable: 
Set  up  the  level  on  the  summit  of  the  hill  at  Gr,  and  fix  the  vane 
on  the  leveling  staff  at  an  elevation  of  four  feet  in  addition  to 
the  height  of  the  telescope  above  the  ground.  Then  direct  an 
assistant  to  carry  the  leveling  staff,  holding  it  in  a  vertical  posi- 
tion, toward  K,  till  he  arrives  at  a  point,  as  a^  where  the  vane 
appears  to  coincide  with  the  cross  wires  of  the  telescope. 
This  will  determine  one  point  of  the  curve  line  four  feet  be- 
low Gr.  The  assistant  may  then  proceed  to  the  line  GrB,  and 
afterward  to  GrL,  moving  backward  or  forward  in  each  of  those 
directions  till  he  finds  points,  as  d  and  g*,  at  which  the  vane 
coincides  with  the  cross  wires  of  the  telescope.  The  horizontal 
distance  between  Gr  and  a,  Gr  and  d^  G-  and  g*,  must  then  be 
measured. 

If  the  leveling  staff  is  sufficiently  long,  the  vane  may  be 
fixed  on  it  at  the  height  of  eight  feet,  in  addition  to  the  height 
of  the  telescope  at  G- ;  and  the  assistant,  placing  himself  in 
the  directions  GrK,  GrB,  GrL,  must  move  till  the  vane  appears 
to  coincide  with  the  cross  wires  as  before.  The  horizontal  dis- 
tances ab,  de^  gh^  must  then  be  measured,  and  stakes  driven 
into  the  ground  at  Z>,  e,  and  h. 

The  level  must  now  be  removed  to  b  ;  and  the  vane  being 
fixed  on  the  staff  at  a  height  equal  to  four  feet,  together  with 
the  height  of  the  instrument  from  the  ground  at  ^,  the  as- 
sistant must  proceed  in  the  direction  6K,  and  stop  at  c  whcF 


126 


Trigonometry. 


the  vane  coincides  with  the  cross  wires ;  then  the  horizontai 

distance  of  c  from  A.      ^^^^ E j^ 

b  must  be  meas- 


ured. 


Miirai 


In  a  like 
manner,  the  op- 
erations may  be 
continued  from  b 
or  c  as  far  as  nec- 
essary toward  K ; 
then,  commenc- 
ing at  e,  and  aft- 
erward at  A,  they 
may  l)e  continur 

ed  in   the    same  C  ]?  D 

way  toward  B  and  L  respectively.  The  angles  whiqh  the  di- 
rections GrK,  GB,  G-L  make  with  the  magnetic  meridian  being 
found  with  the  compass,  these  directions  may  be  represented 
on  paper.  Then  the  measured  distances  Gra,  ab,  &c. ;  Grd^  de^ 
&c. ;  G/>*,  gh,  &c.,  being  set  off  on  those  lines  of  direction, 
ourves  drawn  through  a,  d,  g  ;  b,  e,  h  ;  c,  /,  /(?,  &c.,  will  show 
the  contour  of  the  hill. 

The  map  is  shaded  so  as  to  indicate  the  hills  and  slopes  by 
drawing  fine  lines,  as  in  the  figure,  perpendicular  to  the  hori- 
zontal curves. 

(176.)  Another  method,  which  may  often  be  more  conven- 
ient than  either  of  the  preceding,  is  as  follows :  From  the  sum- 
mit of  the  hill  measure  any  line,  as  GK,  and  at  convenient 
points  of  this  line  let  stakes  be  driven,  and  their  distances  from 
G  be  carefully  measured.  Then  determine  the  difference  of 
level  of  all  these  points  ;  and  if  the  assumed  points  do  not  fall 
upon  the  horizontal  curves  which  are  required  to  be  delineated, 
we  may,  by  supposing  the  slope  to  be  uniform  from  one  stake 
to  another,  compute  by  a  proportion  the  points  where  the  hori- 
zontal curves  for  intervals  of  four  feet  intersect  the  line  GK. 
The  same  may  be  done  for  the  lines  GB  and  GL,  and  for  other 
lines,  if  they  should  be  thought  necessary. 

(177.)  If  the  surface  of  the  ground  is  gently  undulating,  it 
may  be  more  convenient  to  run  across  the  tract  a  number  of 
lines  parallel  to  one  another.     Drive  stakes  at  each  extremity 


Surveying.  127 

t)t  these  lines,  and  also  at  all  the  points  along  them  where 
there  is  any  material  change  in  the  inclination  of  the  ground, 
and  find  the  difference  of  level  between  all  these  stakes,  and 
their  distances  from  each  other.  Then,  if  we  wish  to  draw 
upon  a  map  the  level  lines  at  intervals  of  4,  6,  or  10  feet,  we 
may  compute  in  the  manner  already  explained  the  points 
where  the  horizontal  curves  intersect  each  of  the  parallel  lines. 
The  curve  lines  are  then  to  be  drawn  through  these  points,  ac- 
cording to  the  judgment  of  the  surveyor. 

(178.)  If  it  is  required  to  draw  a  profile  of  the  ground,  \  ^» 
example,  from  (jr  to  K,  draw  a  ^ 

straight  line,  Gr'K,  to  represent  (X^xl 

a  horizontal  line  to  which  the  ^    c^, —  ;       • 

heights  are  referred,  and  set  off  ^,--^-^1       j       j       j 

G-'a',  Gr'b',  G-'c',  &c.,  equal  to     ^^-^^^'^  I      I       i       I , 

the  distances  of  the    stations      -  ^'     ^'    ^'     ^ 

from  the  beginning  of  the  line.  At  the  points  G',  a',  b',  &c., 
erect  perpendiculars,  G-'G,  a'a^  &c.,  and  make  them  equal  to 
the  heights  of  the  respective  stations.  Through  the  tops  of 
these  perpendiculars  draw  the  curved  line  GK,  and  it  will  be 
the  profile  of  the  hill  in  the  direction  of  the  line  GK. 

On  setting  out  Rail-ivay  Curves. 

(179.)  It  is  of  course  desirable  that  the  line  of  a  rail- way 
should  be  perfectly  straight  and  horizontal.  This,  however,  is 
seldom  possible  for  any  great  distance;  and  when  it  becomes 
necessary  to  change  the  direction  of  the  line,  it  should  be  done 
gradually  by  a  curve.  The  curve  almost  universally  employ- 
ed for  this  purpose  is  the  arc  of  a  circle,  and  such  an  arc  may 
be  traced  upon  the  ground  by  either  of  the  following  methods. 

First  Method. — When  the  center  of  the  circle  can  be  seen 
from  every  part  of  the  curve. 

Let  AB,  CD  be  two  straight  portions  of  abroad  which  it  is 
desired  to  connect  by  an  arc  of  a  circle.  Set  up  a  theodolite 
at  B  and  another  at  C,  and  from  each  point  range  a  line  at  right 
angles  to  the  lines  AB  and  CD  respectively ;  and  at  the  inter- 
section of  these  lines,  E,  which  will  be' the  center  of  the  circle, 
erect  a  signal  which  can  be  seen  from  any  point  between  B  and 
C.     Produce  the  lines  AB  and  CD  until  they  meet  in  F.  and 


128 


Trigonometry. 


B  a,   02  ccs  dj.  F 


on  these  lines  drive  stakes  at  equal  distances, 'a,,  a. 

mencing  from  the  points  B  and 

C.     If  r  represents  the  radius    ^^■ 

of  the  circle,  and  d  the  distance 

between  the  points  ^j,  a^,  a^, 

&c.,  then  (Art.  172), 

will  he  the  distance  which  must 
be  set  off  from  the  first  point 
a,,  in  the  direction  a^E,  to  ob- 
tain a  point  of  the  circular  arc. 
Tn  like  manner, 


a,,  corn- 


Vr'+(2^)^-r 
will  be  the  distance  to  be  set  off  from  the  point  a^-,  in  the  di- 
rection 0^2  E  5  ^^dj  generally, 

Vr^+(ndy-r 
will  be  the  distance  to  be  set  off  at  the  «th  points  from  B  and 
C.     For  example,  let  r  be  one  mile,  or  5280  feet,  and  d  equal 
to  100  feet ;  then, 

V5280^+100^-5280=.94  feet, 
will  be  the  distance  a ,  6 , .     In  a  similar  manner,  we  find  at 
a 2,  or  200  feet  from  B,  the  offset  will  be    8.79  feet 
6^3,  or  800  *'  "  "  8.52     " 

^4,  or  400  "  "  "  15.13     " 

.T,,  or  500  "  ''  "  23.62     " 

(180.)  Second  Method. — When  ^  b  a,  p.2  az  g^  05 
the  center  of  the  circle  can  not  be 
seen  from  every  part  of  the  curve, 
the  offsets  may  be  set  off"  perpendic- 
ularly to  the  tangent  BF,  in  which 
case  they  must  be  computed  from 
the  formula 


r-  y/r'-d?. 
For,  in  the  annexed^figure, 


EH=vaE^-aH^ 


that  is,   EH=  ^Jr'-d^. 


And 


a,G=BH=BE-HE=r-  Vr^-rf*. 


Surveying.  12iJ 

If  r=5280  feet,  we  shall  find  the  olTsets  at  intefvals  of  100 
feet  to  1)6  ^1^1=     .95  feet. 

a^h^=  3.79  '' 
a^b^=  8.53  *' 
a, b, =1^11  " 
a, &5  =23.73    « 

For  small  distances,  the  offsets  will  be  given  with  sufficient 
accuracy  by  the  formula 

2r' 
Bee  Art.  172. 

It  is  very  common  for  surveyors,  after  they  have  found  the 
first  point,  Z>  j ,  of  the  curve,  to  join  the  points  B,  6  ^ ,  aud  produce 
the  line  BZ>  ^  to  the  distance  d,  and  from  the  end  of  this  line 
set  off  an  offset  to  determine  the  point  b^ ;  then,  producing  the 
line  6, ^'2 5  S6*  <^ff  3'  third  offset  to  determine  the  point  63,  and 
so  on.  The  objection  to  this  method  is,  that  any  error  com- 
mitted in  setting  out  one  of  the  points  of  the  curve  will  occa- 
sion an  error  in  every  succeeding  one.  Whenever  this  method, 
therefore,  is  employed,  it  should  be  checked  by  determining  the 
position  of  every  fourth  or  fifth  point  by  independent  compu- 
tation and  measurement. 

(181.)  Third  Method. — Where  the  radius  of  the  curve  is 
fTDP.ll,  place  a  theodolite  at  B,  and  point  its  telescope  toward 


C.  Place  another  theodolite  at  C,  and  point  its  telescope  to- 
ward E,  the  point  of  intersection  of  the  lines  AB,  CD  produced. 
Then,  if  the  former  be  moved  through  any  number  of  degrees 
toward  a,,  and  the  latter  the  same  number  of  degrees  toward 
a  ^ ,  the  point  a ,  will  be  a  point  of  the  curve,  for  the  angle 
Ba ,  C  will  be  equal  to  BCD  (Geom.,  Prop.  16,  B.  III.).  In  the 
same  manner,  a^^a^^  &c.,  any  number  of  points  of  the  curve, 
may  be  deV.rmined.     It  will  be  most  convenient  to  miove  the 

I 


130  Trigonometry. 

theodolites  feaoh  time  through  an  even  numher  of  degrees,  for 
example,  an  arc  of  two  degrees,  and  a  stake  must  he  driven  at 
each  of  the  points  of  intersection  a^,  a 2,  a^,  &c.  The  ac- 
curacy of  this  method  is  independent  of  any  undulations  in  the 
surface  of  the  ground,  so  that  in  a  hilly  country  this  method 
may  he  preferahle  to  any  other. 


When  the  position  of  one  end  of  the  curve  is  not  ahsolutely 
determined,  the  engineer  may  proceed  more  rapidly.  Suppose 
it  is  required  to  trace  an  arc  of  a  circle  having  a  curvature  of 
two  degrees  for  a  hundred  feet. 

Place  a  theodolite  at  C,  the  point  where  the  curve  commen- 
ces, and  lay  off  from  the  line  CE,  toward  B,  an  angle  of  two 
degrees,  and  in  the  direction  of  the  axis  of  the  instrument  set 
off  a  distance  of  100  feet,  which  will  give  the  first  point  a^  of 
the  curve.  Next  lay  off  from  CE  an  angle  of  four  degrees,  and 
from  a  ^  set  off  a  distance  of  100  feet,  and  the  point  where  this 
line  cuts  the  axis  of  the  instrument  produced  will  he  the  sec- 
ond point  a^.  In  the  same  manner,  lay  off  from  CE  an  angle 
of  six  degrees,  and  from  a^  set  off  a  distance  of  100  feet,  and 
the  point  where  it  cuts  the  axis  of  the  instrument  produced 
will  he  the  thir4  point  a 3 .  All  the  points  a^,  a^,  a^,  etc.,  thus 
determined  lie  in  the  circumference  of  a  circle  (Geom.,  Prop. 
15,  B.  III.).  Circles  thus  drawn  are  generally  made  with  a 
curvature  of  one  or  two  degrees,  or  some  convenient  fraction 
of  a  degree,  for  every  hundred  feet.  This  method  is  very  ex 
tensively  practiced  in  the  United  States. 

Surveying  Harbors. 
(182.)  In  surveying  a  harhor,  it  is  necessary  to  determine 
the  position  of  the  most  conspicuous  ohjects,  to  trace  the  out- 
line of  the  shore,  and  discover  the  depth  of  water  in  the  neigh- 
borhood  of  the  channel.     A  smooth,  level  piece  of  ground  is 


Surveying. 


131 


chosen,  on  which  a  base  line  of  considerable  length  is  meas- 
ured, and  Station  staves  are  fixed  at  its  extremities.  We  also 
erect  station  staves  on  all  the  prominent  points  to  be  surveyed, 
forming  a  series  of  triangles  covering  the  entire  surface  of  the 
harbor.  The  angles  of  these  triangles  are  now  measured  with 
a  theodolite,  and  their  sides  computed.  After  the  principal 
P'  ints  have  been  determined,  subordinate  points  may  be  ascer- 
t  ined  by  the  compass  or  plane  table. 

Let  the  following  figure  be  a  map  of  a  harbor  to  be  survey- 


ed. We  select  the  most  favorable  position  for  a  base  line, 
which  is  found  to  be  on  the  right  of  the  harbor,  from  A  to  B. 
We  also  erect  station  flags  at  the  points  C,  D,  E,  F,  and  Gr. 
Having  carefully  measured  the  base  line  AB,  we  measure  the 
three  angles  of  the  triangle  ABC,  which  enables  us  to  compute 
the  remaining  sides.  We  then  measure  the  three  angles  of  the 
triangle  ACD,  and  by  means  of  the  side  AC,  just  computed, 
we  are  enabled  to  compute  AD  and  CD.  We  then  measure 
the  three  angles  of  the  triangle  CDF,  ftnd  by  means  of  the  side 
CD,  just  found,  we  are  enabled  to  compute  CF  and  DF.  Pro- 
ceeding in  the  same  manner  with  the  triangles  CEF,  DFG-,  we 
are  enabled,  after  measuring  the  angles,  to  compute  the  sides. 
(183.)  Having  determined  the  main  points  of  the  harbor,  we 
may  proceed  to  a  more  detailed  survey  by  means  of  the  chain 


132 


Trigonometry. 


and  compass.  If  it  is  required  to  trace  the  shore,  HCK,  w<j 
commence  at  H,  and  observe  the  bearings  with  the  compase, 
and  measure  the  distances  with  the  chain.  Where  the  shore 
is  undulating,  it  is  most  convenient  to  run  a  straight  line  foi 
a  considerable  distance,  and  at  frequent  intervals  measure  off- 
sets to  the  shore. 

When  a  great  many  objects  are  to  be  represented  upon  9 
map,  the  most  convenient  instrument  is 

The  Plane  Table. 

(184.)  The  plane  table 
is  a  board  about  sixteen 
inches  square,  designed 
to  receive  a  sheet  of  draw- 
ing paper,  and  has  two 
plates  of  brass  upon  op- 
posite sides,  confined  by- 
screws,  for  stretching  and 
retaining  the  paper  upon 
the  board.  The  margin 
of  the  board  is  divided  to 
360  degrees  from  a  cen- 
ter C,  in  the  middle  of 
the  board,  and  these  are 
subdivided  as  minutely  as  the  size  of  the  table  will  admit.  On 
one  side  of  the  board  there  is  usually  a  diagonal  scale  of  equal 
parts.  A  compass  box  is  sometimes  attached,  which  renders 
the  plane  table  capable  of  answering  the  purpose  of  a  survey- 
or^s  compass. 

The  ruler.  A,  is  made  of  brass,  as  long  as  the  diagonal  of 
the  table,  and  about  two  inches  broad.  A  perpendicular  sight- 
vane,  B,  B,  is  fixed  to  each  extremity  of  the  ruler,  and  the  eye 
looking  through  one  of  them,  the  vertical  thread  in  the  other  is 
made  to  bisect  any  required  distant  object. 

To  the  under  side  of  the  table,  a  center  is  attached  with  a 
ball  and  socket,  or  parallel  plate  screws,  like  those  of  the  the- 
odolite, by  which  it  can  be  placed  upon  a  staff-head ;  and  the 
table  may  be  made  horizontal  by  means  of  a  detached  spirit 
level 


Surveying.  133 

(185.)  To  prepare  the  table  for  use,  it  must  be  covered  with 
drawing  paper.  Then  set  up  the  instrument  at  one  of  the 
stations,  for  example,  B  (see  fig.  on  p.  131),  and  fix  a  needle  in 
the  table  at  the  point  on  the  paper  representing  that  station, 
and  place  the  edge  of  the  ruler  against  the  needle.  Then  di- 
rect the  sights  to  the  station  A,  and  by  the  side  of  the  ruler 
draw  a  line  upon  the  paper  to  represent  the  direction  of  AB. 
Then,  with  a  pair  of  dividers,  take  from  the  scale  a  certain 
number  of  equal  parts  to  represent  the  base,  and  lay  off  this 
distance  on  the  base  line.  Having  drawn  the  base  line,  move 
the  ruler  around  the  needle,  direct  the  sights  to  any  object, 
as  L,  and  keeping  it  there,  draw  a  line  along  the  edge  of  the  ru- 
ler. Then  direct  the  sights  in  the  same  manner  to  any  other 
objects  which  are  required  to  be  sketched,  drawing  lines  in  thei? 
respective  directions,  taking  care  that  the  table  remains  steady 
during  the  operation. 

Now  remove  the  instrument  to  the  other  extremity  of  the 
base  A,  and  place  the  point  of  the  paper  corresponding  to  that 
extremity  directly  over  it.  Place  the  edge  of  the  ruler  on  the 
base  line,  and  turn  the  table  about  till  the  sights  are  directed 
to  the  station  B.  Then  placing  the  edge  of  the  ruler  against 
the  needle,  direct  the  sights  in  succession  to  all  the  objects  ob- 
served from  the  other  station,  drawing  lines  from  the  point  A 
in  their  several  directions.  The  intersections  of  these  lines 
with  those  drawn  from  the  point  B  will  determine  the  posi- 
tions of  the  several  objects  on  the  map. 

In  this  manner  the  plane  table  may  be  employed  for  filling 
in  the  details  of  a  map  ;  setting  it  up  at  the  most  remarkable 
spots,  and  sketching  by  the  eye  what  is  not  necessary  should 
be  more  particularly  determined,  the  paper  will  gradually  be- 
come a  representation  of  the  country  to  be  surveyed. 

To  determine  the  Depth  of  Water. 
(186.)  Let  signals  be  established  on  the  principal  shoals  and 
along  the  edges  of  the  channel,  by  erecting  poles  or  anchoring 
buoys,  and  let  their  bearings  be  observed  from  two  stations  of 
the  survey.  Then  in  each  triangle  there  v/ill  be  known  one 
side  and  the  angles,  from  which  the  other  sides  may  be  com- 
put<^d,  and  their  positions  thus  become  known.     Then  asoer- 


134  Trigonometry. 

tain  the  precise  depth  of  water  at  each  of  the  buoys,  and  pro- 
ceed  in  this  manner  to  determine  as  many  points  as  may  be 
thought  necessary. 

If  an  observer  is  stationed  with  a  theodolite  at  each  extremi- 
ty of  the  base  line,  we  may  dispense  with  the  erection  of  per- 
manent marks  upon  the  water.  One  observer  in  a  boat  may 
make  a  sounding  for  the  depth  of  water,  giving  a  signal  at  the 
same  instant  to  two  observers  at  the  extremities  of  the  base 
line.  The  dk^ection  of  the  boat  being  observed  at  that  instant 
from  two  stations,  the  precise  place  of  the  boat  can  be  com- 
puted. In  this  way  soundings  may  be  made  with  great  ex- 
pedition. X 

There  is  also  another  method,  still  more  expeditious,  which 
may  afford  results  sufficiently  precise  in  some  cases.  Let  a 
boat  be  rowed  uniformly  across  the  harbor  from  one  station  to 
another,  for  example,  from  D  to  Gr  (see  fig.  on  p.  131),  and  let 
a  series  of  soundings  be  made  as  rapidly  as  possible,  and  the 
instant  of  each  sounding  be  recorded.  Then,  knowing  the  en- 
tire length  of  the  line  DGr,  and  the  time  of  rowing  over  it,  we 
may  find  by  proportion  the  approximate  position  of  the  boat  at 
each  sounding. 

If  the  soundings  are  made  in  tide  waters,  the  times  of  high 
water  should  be  observed,  and  the  time  of  each  sounding  be 
recorded,  so  that  the  depth  of  water  at  high  or  low  tide  may 
be  computed.  In  the  maps  of  the  United  States  Coast  Survey, 
the  soundings  are  all  reduced  to  low-water  mark,  and  the  num- 
ber of  feet  which  the  tide  rises  or  falls  is  noted  upon  the  map 

(187.)  The  results  of  the  soundings  may  be  delineated  upon 
a  map  in  the  same  manner  as  the  observations  of  level  on  page 
124.  We  draw  lines  joining  all  those  points  where  the  depth 
of  water  is  the  same,  for  example,  20  feet.  Such  a  line  is  seen 
to  be  an  undulating  line  running  in  the  direction  from  E  to  G. 
We  draw  another  line  connecting  all  those  points  where  the 
depth  of  water  is  40  feet.  This  line  runs  somewhat  to  the 
east  of  the  former  line,  but  nearly  parallel  with  it.  We  draw 
other  lines  for  depths  of  60  feet,  &c.  The  lines  being  thus 
drawn,  a  mere  glance  at  the  map  will  show  nearly  the  depth 
of  water  at  any  point  of  the  harbor. 


BOOK  V. 

NAVIGATION. 

(I'yiX)  Navigation  is  the  art  of  conducting  a  ship  at  sea 
from  <  ne  port  to  another. 

Tliyre  are  two  methods  of  determining  the  situation  of  a 
vessel  at  sea.  The  one  consists  in  finding  by  astronomical  oh- 
servations  her  latitude  and  longitude ;  the  other  consists  in 
measuring  the  ship's  course,  and  her  progress  every  day  from 
the  time  of  her  leaving  port,  from  which  her  place  may  be 
computed  by  trigonometry.  The  latter  method  is  the  one  to 
be  now  considered. 

(189.)  The  figure  of  the  earth  is  nearly  that  of  a  sphere,  and 
in  navigation  it  is  considered  perfectly  spherical.  The  earth's 
axis  is  the  diameter  around  which  it  revolves  once  a  day. 
The  extremities  of  this  axis  are  the  terrestrial  poles  ;  one  is 
called  the  north  pole,  and  the  other  the  south  pole. 

The  equator  is  a  great  circle  perpendicular  to  the  earth's 
axis. 

Meridians  are  great  circles  passing  through  the  poles  of  the 
earth.    Every  place  on  the  earth's  surface  has  its  own  meridian. 

(190.)  The  longitude  of  any  place  is  the  arc  of  the  equator 
intercepted  between  the  meridian  of  that  place  and  some  as- 
sumed meridian  to  which  all  others  are  referred.  In  most 
countries  of  Europe,  that  has  been  taken  as  the  standard  me- 
ridian which  passes  through  their  principal  observatory.  The 
English  reckon  longitude  from  the  Observatory  of  Grreenwich ; 
and  in  the  United  States,  we  have  usually  adhered  to  the  En- 
glish custom,  though  we  believe  the  time  has  come  when  longi- 
tude should  be  reckoned  from  the  Observatory  of  Washington. 

Longitude  is  usually  reckoned  east  and  west  of  the  first  me- 
ridian, from  0^  to  180°. 

The  difference  of  longitude  of  two  places  is  the  arc  of  the 
equator  included  between  their  meridians.     It  is  equal  to  the 


136  Trigonometry. 

difference  of  their  longitudes  if  they  are  on  the  same  side  of 
the  first  meridian,  and  to  the  sum  of  their  longitudes  if  on  op. 
posite  sides. 

(191.)  The  latitude  of  a  place  is  the  arc  of  the  meridian  pass- 
ing through  the  place,  which  is  comprehended  between  that 
place  and  the  equator. 

Latitude  is  reckoned  north  and  south  of  the  equator,  from 
0°  to  90°. 

Parallels  of  latitude  are  the  circumferences  of  small  circles 
parallel  to  the  equator. 

The  difference  of  latitude  of  two  places  is  the  arc  of  a  me- 
ridian included  between  the  parallels  of  latitude  passing 
through  those  places.  It  is  equal  to  the  difference  of  their 
latitudes  if  they  are  on  the  same  side  of  the  equator,  and  to 
the  sum  of  their  latitudes  if  on  opposite  sides. 

The  distance  is  the  length  of  the  line  which  a  vessel  de- 
scribes in  a  given  time. 

The  departure  of  two  places  is  the  distance  of  either  place 
from  the  meridian  of  the  other.  If  the  two  places  are  on 
the  same  parallel,  the  departure  is  the  distance  between  the 
places.  Otherwise,  we  divide  the  distance  AB  into  portion? 
A6,  be,  cd,  &o.,  so  small 
that  the  curvature  of  the 
earth  may  be  neglected. 
Through  these  points 
we  draw  the  meridians 
P^,  Pc,  &c.,  and  the  par- 
allels be,  cf,  &c.  Then 
the  departure  for  Ab  is 
eb,  for  be  it  is  fc  ;  and 
the  whole  departure  from  A  to  B  is  eb+fc+g-d+^B ;  that  is, 
the  sum  of  the  departures  corresponding  to  the  small  portions 
into  which  the  distance  is  divided. 

Distance,  departure,  and  difference  of  latitude  are  measured 
in  nautical  miles,  one  of  which  is  the  60th  part  of  a  degree  at 
the  equator.  A  nautical  mile  is  nearly  one  sixth  greater  than 
an  English  statute  mile. 

The  course  of  a  ship  is  the  angle  which  the  ship's  path  makes 
with  the  meridian.     A  ship  is  said  to  continue  on  the  same 


Navigation. 


13" 


course  when  she  cuts  every  meridian  which  she  crosses  at  the 
same  angle.  The  path  thus  described  is  not  a  straight  line, 
but  a  curve  called  a  rhumb-line. 

The  course  of  a  ship  is  given  by  the  mariner's  compass. 

(192.)  The  mariner^s  compass  consists  of  a  circular  piece  of 
paper,  called  a  card,  attached  to  a  magnetic  needle,  which  is 
balanced  on  a  pin  so  as  to  move  freely  in  any  direction.  Di 
rectly  over  the  needle,  a  line  is  drawn  on  the  card,  one  end  ol 
which  is  marked  N,  and  the  other  S.  The  circumference  is 
divided  into  thirty-two  equal  parts  called  rhumbs  or  points, 
each  point  being  subdivided  into  four  equal  parts  called  quarter 
points. 

The  points  of  the  compass  are  designated  as  follows,  begin- 
ning at  north  and  go- 
ing east:  north,  north 
by  east,  north-north- 
east, northeast  by 
north,  northeast,  and 
so  on,  as  shown  in  the 
annexed  figure. 

The  interval  be- 
tween two  adjacent 
points  is  11°  15', 
which  is  the  eighth 
part  of  a  quadrant. 
On  the  inside  of  the 
compass-box  a  black 
line  is  drawn  perpen- 
dicular to  the  horizon,  and  the  compass  should  be  so  placea 
that  a  line  drawn  from  this  mark  through  the  center  of  the 
card  may  be  parallel  to  the  keel  of  the  ship.  The  part  of  the 
card  which  coincides  with  this  mark  will  then  show  the  point 
of  the  compass  to  which  the  keel  is  directed.  The  compass  is 
suspended  in  its  box  in  such  a  manner  as  to  maintain  a  hori- 
zontal position,  notwithstanding  the  motion  of  the  ship. 

The  following  table  shows  the  number  of  degrees  and  min- 
ates  corresponding  to  each  point  of  the  compass  : 


138 


Trigonometry 


North. 

Pts. 

Pt8. 

South. 

N.by  E. 

N.byW. 

1 

11°  15' 

1 

S.byE. 

S.byW. 

N.N.E. 

N.N.W. 

2 

22°  30' 

2 

S.S.E. 

s.s.w. 

N.E.byK 

N.W.byN. 

3 

33°  45' 

3 

S.E.byS. 

S.W.byS.J 

N.E. 

N.W. 

4 

45°    0' 

4 

S.E. 

s.w. 

N.E.byE. 

N.W.byW. 

5 

56°  15' 

5 

S.E.byE. 

S.W.byW. 

E.N.E. 

W.N.W. 

6 

67°  30' 

6 

E.S.E. 

W.S.W. 

E.byN. 

W.byN. 

7 

78°  45' 

7 

E.byS. 

W.  by  S. 

East. 

West. 

8 

90°    0' 

8 

East. 

West. 

(193.)  The  ship's  rate  of  sailing  is  measured  by  a  log-line. 
The  log-line  is  a  cord  about  300  yards  long,  which  is  wound 


round  a  reel,  one  end  being  attached  to  a  piece  of  thin  board 
called  a  log.  This  board  is  in  the  form  of  a  sector  of  a  circle, 
the  arc  of  which  is  loaded  with  lead  sufficient  to  give  the  board 
a  vertical  position  when  thrown  upon  the  water.  This  is  de- 
signed to  prevent  the  log  from  being  drawn  along  after  the 
vessel  while  the  line  is  running  off  the  reel. 

The  time  is  measured  by  a  sand-glass,  through  which  'the 
sand  passes  in  half  a  minute^  or  the  120th  part  of 
an  hour. 

The  log-line  is  divided  into  equal  parts  called 
knots,  each  of  which  is  50  feet,  or  the  120th  part 
of  a  nautical  mile.  Now,  since  a  knot  has  the 
same  ratio  to  a  nautical  mile  that  half  a  minute 
has  to  an  hour,  it  follows,  that  if  the  motion  of  a 
ship  is  uniform,  she  sails  as  many  miles  in  an  hour 
as  she  does  knots  in  half  a  minute.  If,  then,  seven  knots  are 
observed  to  run  off  in  half  a  minute,  the  ship  is  sailing  at  the 
rate  of  seven  miles  an  hour. 


PLANE  SAILING. 
(194.)  Plane  sailing'  is  the  method  of  calculating  a  ship's 
place  at  sea  by  means  of  the  properties  of  a  plane  triangle. 
The  particulars  which  are  given  or  required  are  four,  viz.,  the 


Navigation. 


139 


Hence  the  difference  of  lati- 


distance,  course,  difference  of  latitude,  and  departure.     Of 
these,  any  two  being  given,  the  others  may  be  found. 

Let  the  figure  EPQ,  represent  a  portion  of  the  earth's  sur- 
face, P  the  pole,  and  EQ, 
the  equator.  Let  AB 
be  a  rhumb-line,  or  the 
track  described  by  a  ship 
in  sailing  from  A  to  B 
on  a  uniform  course. 
Let  the  whole  distance 
be  divided  into  portions 
A6,  6c,  &c.,  so  small 
that  the  curvature  of  the  earth  may  be  neglected.  Through 
the  points  of  division  draw  the  meridians  P6,  Pc,  &c.,  and  the 
parallels  eb^  fc,  &c.  Then,  since  the  course  is  every  where 
the  same,  each  of  the  angles  ekb,  fbc^  &c.,  is  equal  to  the 
course.  The  distances  Ae,  bf,  &c.,  are  the  differences  of  lati- 
tude of  A  and  6,  b  and  c,  &o.  Also,  eb^  fc^  &c.,  are  the  de- 
partures for  the  same  distances, 
tude  from  A  to  B  is  equal  to 

ke+bf^cg-Vdh, 
and  the  departure  is  equal  to 
eb-\-fc-\-gd-\-hB. 

Construct  the  triangle  A'B'C  so  that  K'b'e' 
shall  be  equal  to  kbe^  b'c'f  shall  be  equal  to 
bcf,  c'd'g'  equal  to  cdg,  and  d'Wh'  equal  to 
dQh.     Then  A'B'  represents  the  distance  sail- 
ed, B'A'C  the  course,  A'C  the  difference  of  latitude,  and  B'C 
the  departure ;  that  is,  the  distance,  dif- 
ference of  latitude,  and  departure  are  cor- 
rectly represented  by  the  hypothenuse  and 
sides  of  a  right-angled  triangle,  of  which 
the  angle  oppoiite  to  the  departure  is  the 
course.     Of  these  four  quantities,  any  two 
being  given,  the  others  may  be  found. 

Plane  sailing  does  not  assume  the  earth's 
surface  to  be  a  plane,  and  does  not  involve 
any  error  even  in  great  distances. 


Departure 


140  Trigonometry. 

Examples.  * 

1.  A  ship  sails  from  Vera  Cruz  N.E.  by  N".  74  miles.  Re- 
quired  her  departure  and  difference  of  latitude. 

According  to  the  principles  of  right-angled  triangles,  Art.  44. 
Radius  :  distance  : :  sin.  course  :  departure. 

: :  COS.  course  :  diff.  latitude. 
The  course  is  three  points,  or  33°  45' ;  hence  we  oh  tain 

Departure      =41.11  miles. 
Biff.  latitude= 61.53  miles. 

2.  A  ship  sails  from  Sandy  Hook,  latitude  40°  28'  N.,  upon 
a  course  E.S.E.,  till  she  makes  a  departure  of  500  miles.  What 
distance  has  she  sailed,  and  at  what  latitude  has  she  arrived  ? 

By  Trigonometry,  Art.  44, 

Sin.  course  :  departure  : :  radius  :  distance^ 

: :  COS.  course  :  diff.  latitude* 
Ans.  Distance        ==541.20  miles. 

Diff.  latitude=207.11  miles,  or  3°  27'. 
Hence  the  latitude  at  which  she  has  arrived  is  37°  1'  N. 

3.  The  hearing  of  Sandy  Hook  from  Bermuda  is  N.  42°  56' 
W.,  and  the  difference  of  latitude  486  milos.  Required  the 
distance  and  departure. 

By  Trigonometry,  Art.  46, 

Radius  :  diff.  latitude  : :  tang,  course  :  departure^ 
: :  sec.  course  :  distance. 
Ans.  Distance   =663.8  miles. 
%  Departure =452.1  miles. 

4.  A  ship  sails  from  Bermuda,  latitude  32°  22'  N.,  a  distance 
of  666  miles,  upon  a  course  between  north  and  east,  until  she 
finds  her  departure  444  miles.  What  course  has  she  sailed, 
and  what  is  her  latitude  ? 

By  Trigonometry,  Art.  44,  • 

Distance  :  radius  : :  departure  :  sin.  course, 
,  .  Radius  :  distance  : :  cos.  course  :  diff.  latitude. 

Ans.  Latitude=      40°  38'  N. 
Course    =N.  41°  49'  E. 

5.  The  distance  from  Vera  Cruz,  latitude  19°  12'  N.,  to  Pen- 
Bacola,  latitude  30°  19'  N.,  is  820  miles.  Required  the  hear- 
ing and  departure. 


Navigation.  141 

By  Trigonometry,  Art.  45, 

Distance  :  radius  : :  diff.  latitude  :  cos.  course, 
Radius  :  distance  : :  sin.  course  :  departure. 

Ans.  Bearing    =N.  35°  34'  E. 
Departure =476.95  miles. 
6.  A  ship  sails  from  Sandy  Hook  upon  a  course  between 
south  and  east  to  the  parallel  of  35°,  when  her  departure  was 
300  miles.     Required  her  course  and  distance. 
By  Trigonometry,  Art.  47, 

Diff.  latitude  :  radius  : :  departure  :  tang,  course^ 
B/idius  :  diff.  latitude  : :  sec.  course  :  distance. 

Ans.  Course  S.  42°  27'  E. 
Distance  444.5  miles. 

TRAVERSE  SAILING. 

{195.)  A  traverse  is  the  irregular  path  of  a  ship  when  sail- 
ing on  different  courses. 

The  object  of  traverse  sailing"  is  to  reduce  a  traverse  to  a 
single  course,  when  the  distances  sailed  are  so  small  that  the 
curvature  of  the  earth  mdy  be  neglected.  When  a  ship  sails 
on  different  courses,  the  difference  of  latitude  is  equal  to  the 
difference  between  the  sum  of  the  northings  and  the  sum  of  the 
southings ;  and,  neglecting  the  earth's  curvature,  the  departure 
is  equal  to  the  difference  between  the  sum  of  the  eastings  and 
the  sum  of  the  westings.  If,  then,  the  difference  of  latitude  and 
the  departure  for  each  course  be  taken  from  the  traverse  table, 
and  arranged  in  appropriate  columns,  the  difference  of  latitude 
for  the  whole  time  may  be  obtained  exactly,  and  the  departure 
nearly,  by  addition  and  subtraction;  and  the  corresponding 
distance  and  course  may  be  determined  as  in  plane  sailing. 

Examples. 
I.  A  ship  sails  on  the  following  successive  tracks : 

1.  N.E.      23  miles. 

2.  E.S.E.    45      " 

3.  E.byN.  34      " 

4.  North     29      " 

5.  N.  by  W.  31      " 

6.  N.N.E.    17      *' 

Find  the  course  and  distance  for  the  whole  traverse. 


142 


Trigonometry. 


We  form  a  table  as  below,  entering  the  courses  from  th« 
table  of  rhumbs,  page  138,  and  then  enter  the  latitudes  and 
departures  taken  from  the  traverse  table. 

Traverse  Table. 


No. 

Course. 

Distance. 

N. 

s. 

E. 

w. 

1 

N.  45°  E. 

23 

16.26 

16.26 

2 

S.  67°  30'  E. 

45 

17.22 

41.57 

3 

N.  78°  45'  E. 

34 

6.63 

83.35 

4 

North. 

29 

29.00 

:; 

5 

N.  11°  15'  W. 

31 

30.40 

6.05 

6 

N.  22°  30'  E. 

17 

15.71 

6.51 

Sum  of  columns 


98.00 
17.22 


17.22 


97.69 
6.05 


6.05 


Diff.  latitude     .     .     .     .=80.78  N.    Dep.=91.64  E. 

Hence  the  course  is  found  by  plane  sailing  N.  48°  36'  E., 
and  the  distance  =122.2  miles. 

The  proportions  are 

Diff.  latitude  :  radius  : :  departure  :  tang',  course, 
Radius  :  diff.  latitude  : :  sec.  course  :  distance. 

2.  A  ship  leaving  Sandy  Hook  makes  the  following  courses 
and  distances : 

1.  S.E. 

2.  E.S.E. 

3.  East 

4.  E.byS. 

5.  South 

6.  S.byE. 
Required  her  latitude,  the   distance   made, 
course. 

Ans.  Latitude=38°  V  N. 
Distance =193.7  miles. 
Course    =S.  40°  47'  E. 

3.  A  ship  from  Pensacola,  latitude  30°  19',  sails  on  the  foL 
lowing  successive  courses : 

1.  South      48  miles. 

2.  S.S.W.      23      " 

3.  S.W.       32      " 


25  miles 

32 

17 

51 

45 

63 

and  the  direct 


N 


AVIGATION. 


143 


4.  S.W.  by  S.  76  miies. 

5.  West       17      " 

6.  W.S.W.     54     " 
Required  her  latitude,  direct  course,  and  distance. 

Ans.  Latitude^     27°  23'  N. 
Course    =S.38°39'W. 
Distance  =225.0  miles. 
4.  A  ship  from  Bermuda,  latitude  32°  22',  sails  on  the  fol- 
lowing successive  courses : 


66  miles. 
14      " 
45      " 
21      " 
32      " 


1.  N.E. 

2.  KN.E. 

3.  N.E.byE. 

4.  East 

5.  E.byN. 
Required  her  latitude,  direct  course,  and  distance. 

Ans.  Latitude^  33°  53'  N. 
Course  =N.  57°  22'  E. 
Distance^  168.4  miles. 

(196.)  When  the  water  through  which  a  ship  is  moving  has 
a  progressive  motion,  the  ship's  progress  is  affected  in  the  same 
manner  as  if  she  had  sailed  in  still  water,  with  an  additional 
course  and  distance  equal  to  the  direction  and  motion  of  the 
current. 

Ex.  5.  If  a  ship  sail  125  miles  N.N.E.  in  a  current  which 
sets  W.  by  N.  32  miles  in  the  same  time,  required  her  true 
course  and  distance. 

Form  a  traverse  table  containing  the  course  sailed  by  the 
ship  and  the  progress  of  the  current,  and  find  the  difference 
of  latitude  and  departure.  The  resulting  course  and  distance 
is  found  as  in  the  preceding  examples. 


Traverse  Table. 

Courses. 

Distance. 

N. 

E. 

w. 

N.  22°  30'  E 

N.  78°  45'  W. 

125 
32 

115.49 
6.24 

47.84 

31.39 

DifF.  latitude 


Departure    . 


121.73 


47.84 
31.39 

=16.45  E. 


31.39 


144  Trigonometry. 

Hence  the  course  is  found  by  plane  sailing  N.  7°  42'  E.,  and* 
the  distance =122.8  miles. 

Ex.  6.  A  ship  sails  S.  by  E.  for  two  hours  at  the  rate  of  9 
miles  an  hour ;  then.  S.  by  W.  for  five  hours  at  the  rate  of  8 
miles  an  hour ;  and  during  the  whole  time  a  current  sets  W. 
by  N.  at  the  rate  of  two  and  a  half  miles  an  hour.  Required 
the  direct  course  and  distance. 

Ans.  The  course  is  S.  21°  51'  W. 
Distance  57.6  miles. 

0 

PARALLEL  SAILING. 
(197.)  Parallel  sailing'  is  when  a  ship  sails  exactly  east  oj 
west,  and  therefore  remains  constantly  on  the  same  parallel 
of  latitude.  In  this  case  the  departure  is  equal  to  the  distance 
sailed,  and  the  difference  of  longitude  may  be  found  by  the  fol- 
lowing 

Theorem. 

The  cosine  of  the  latitude  of  the  parallel  is  to  radius,  as 
the  distance  run  is  to  the  difference  of  longitude. 

Let  P  be  the  pole  of  the  earth,  C  the  center,  AB  a  portion 
of  the  equator,  and  DE  any  parallel  of  lati-  p 
tude ;  then  will  CA  be  the  radius  of  the 
equator,  and  FD  the  radius  of  the  parallel. 
Let  DE  be  the  distance  sailed  by  the  ship 
on  the  parallel  of  latitude,  then  the  difference 
of  longitude  will  be  measured  by  AB,  the 
arc  intercepted  on  the  equator  by  the  merid-  ^\^  ^^ 

ians  passing  through  D  and  E.  -^ 

Since  AB  and  DE  correspond  to  the  equal  angles  ACB, 
DFE,  they  are  similar  arcs,  and  are  to  each  other  as  their 
radii.     Hence 

FD  :  CA  : :  arc  DE  :  arc  AB. 

But  FD  is  the  sine  of  PD,  or  the  cosine  of  AD,  that  is,  the 
cosine  of  the  latitude,  and  CA  is  the  radius  of  the  sphere; 
hence 

Cosine  of  latitude  :  R  :  :  distance  :  diff.  longitude. 

Cor.  Like  portions  of  different  parallels  of  latitude  are  tc 
each  other  as  the  cosines  of  the  latitudes. 


Navigation. 


145 


The  length  of  a  degree  of  longitude  in  different  parallels  may 
De  computed  by  this  theorem.  A  degree  of  longitude  at  the 
equator  being  60  nautical  miles,  a  degree  in  latitude  40®  may 
be  found  by  the  proportion 

R  :  cosine  40°  :  :  60  :  45.96,  the  required  length. 

The  following  table  is  computed  in  the  same  manner. 

(198.)  Table  showing  the  length  of  a  degree  of  longittide 
for  each  degree  of  latitude. 


|Lat. 

Miles. 

Lat.|    Miles. 

Lat.j    Miles. 

Lat. 

Miles. 

LaL|    Miles. 

Lat. 

Miles.   1 

1 

59.99 

16157.68 

31 

51.43 

46 

41.68 

61 

29.09 

76 

14.52 

2 

59.96 

17 

57.38 

32 

50.88 

47 

40.92 

62 

28.17 

77 

13.50 

3 

59.92 

18|  57.06 

33 

50.32 

48 

40.15 

63 

27.24 

78 

12.47 

i  4 

59.85 

19:56.73 

34 

49.74 

49 

39.36 

64 

26.30 

79 

11.45 

1  5 

59.77 

20 

56.38 

35 

49.15 

50 

38.57 

Q6 

25.36 

80 

10.42 

6 

59.67 

21 

56.01 

36 

48.54 

51 

37.76 

m 

24.40 

81 

9.39 

7 

59.55 

22 

55.63 

37 

47.92 

52 

36.94 

67 

23.44 

82 

8.35 

8 

59.42 

23 

55.23 

38 

47.28 

53 

36.11 

68 

22.48 

83 

7.31 

9 

59.26 

24 

54.81 

39 

46.63 

54 

35.27 

69 

21.50 

84 

6.27 

10 

59.09 

25 

54.38 

40 

45.96 

6^ 

34.41 

70 

20.52 

85 

5.23 

11 

58.90 

26 

53.93 

41 

45.28 

5Q 

33.55 

71 

19.53 

86 

4.19 

12 

58.69 

27 

53.46 

42 

44.59 

57 

32.68 

72 

18.54 

87 

3.14 

13 

58.46 

28 

52.98 

43 

43:88 

58 

31.80 

73 

17.54 

88 

2.09 

14 

58.22 

29 

52.48 

44 

43.16 

59 

30.90 

74 

16.54 

89 

1.05 

15 

57.96 

30 

51.96 

45 

42.43 

60 

30.00 

75 

15.53 

90 

0.00 

B 


Let  ABC  represent  a  right-angled  triangle ;  then,  by  Trig- 
onometry, Art.  41, 

cos.  B  :  R  :  :  AB  :  BC. 
But,  by  the  preceding  Theorem,  we  have 
cos.  lat.  :  R  :  :  depart.  :  diff.  long., 

from  which  we  see  that  if  one  leg  of  a  

right-angled  triangle   represent  the   dis-  ^'^^   "'^^ 

tance  run  on  any  parallel,  and  the  adjacent  acute  angle  be 
made  equal  to  the  degrees  of  latitude  of  that  parallel,  then  the 
hypothenuse  will  represent  the  difference  of  longitude. 

Examples. 
1.  A  ship  sails  from  Sandy  Hook,  latitude  40°  28'  N.,  longi- 
tude 74°  V  W.,  618  miles  due  east.     Required  her  present 
longitude. 

Cos.  40°  28'  :  R  : :  618  ;  812'.3=13°  32',  the  difference  of 
mnffitude. 

K 


146  Trigonometry. 

This,  subtracted  from  74°  1',  leaves  60°  29'  W.,  the  longi- 
tude  required. 

2.  A  ship  in  latitude  40°  sai^s  due  east  through  nine  degrees 
of  longitude.     Required  the  distance  run. 

Ans.  413.66  miles. 

3.  A  ship  having  sailed  on  a  parallel  of  latitude  261  miles, 
finds  her  difference  of  longitude  6°  15'.     What  is  her  latitude  ? 

Ans.  Latitude  45°  54'. 

4.  Two  ships  in  latitude  52°  N.,  distant  from  each  other  95 
miles,  sail  directly  south  until  their  distance  is  150  mile^ 
What  latitude  do  they  arrive  at  ? 

Ans.  Latitude  13°  34' 

MIDDLE  LATITUDE  SAILING. 

(199.)  By  the  method  just  explained  may  be  found  the  lon- 
gitude which  a  ship  makes  while  sailing  on  a  parallel  of  lati- 
tude. When  the  course  is  oblique,  the  departure  may  be  found 
by  plane  sailing,  but  a  difficulty  is  found  in  converting  this 
departure  into  difference  of  longitude. 

If  a  ship  sail  from  A  to  B,  the  departure  is  equal  to  eb+fc 
■^gd+hB,  which  is  less  than  AC,  but 
greater  than  DB.  Navigators  have  as- 
sumed that  the  departure  was  equal  to 
the  distance  between  the  meridians  PA, 
PB,  measured  on  a  parallel  EF,  equidis- 
tant from  A  and  B,  called  the  middle  lati- 
tude. 

The  middle  latitude  is  equal  to  half  the 
sum  of  the  two  extreme  latitudes,  if  both 
are  north  or  both  south ;  but  to  half 'their  difference,  if  one  is 
north  and  the  other  south. 

The  principle  assumed  in  middle  latitude  sailing  is  not  per- 
fectly correct.  For  long  distances  the  error  is  considerable, 
but  the  method  is  rendered  perfectly  accurate  by  applying  to 
the  middle  latitude  a  correction  which  is  given  in  the  accom 
panying  tables,  page  149. 

(200.)  It  has  been  shown  that  when  a  ship  sails  upon  an 
oblique  course,  the  distance,  departure,  and  difference  of  lati 
tude  may  be  represented  by  the  sides  of  a  right-angled  triaD 


Navigation.  147 

gle.  The  difference  of  longitude  is  derived  from  the  departure, 
in  the  sgme  manner  as  in  parallel  sailing,  the  ship  being  sup- 
posed to  sail  on  the  middle  latitude  parallel.  Hence,  if  we 
combine  the  triangle  ABC  for  plane  sailing 
with  the  triangle  BCD  for  parallel  sailing, 
we  shall  obtain  a  triangle  ABD,  by  which 
all  the  cases  of  middle  latitude  sailing  may- 
be solved. 

In  the  triangle  BCD, 

Cos.  CBD  :BC  : :  R  :  BD  ; 
that  is,  cosine  of  middle  latitude  is  to  the 
departure^  as  radius  is  to  the  difference  of 
longitude. 

In  the  triangle  ABD,  since  the  angle  D  is 
the  complement  of  CBD,  which  represents  the  middle  latitude, 
we  have 

Sin.  D  :  AB  : :  sin.  A  :  BD ; 
that  is,  cosine  of  middle  latitude  is  to  the  distance^  as  the  sine 
of  the  course  is  to  the  difference  of  longitude. 

In  the  triangle  ABC,  we  have  the  proportion 
AC  :  BC  :  :  R  :  tang.  A. 

But  we  have  before  had  the  proportion 

Cos.  CBD  :  BC  :  :  R  :  BD. 

The  means  being  the  same  in  these  two  proportions,  we  have 
Cos.  CBD  :  AC  : :  tang.  A  :  BD ; 
that  is,  cosine  of  middle  latitude  is  to  the  difference  of  lati- 
tude^ as  the  tangent  of  the  course  is  to  the  difference  of  lon- 
gitude. 

The  middle  latitude  should  always  be  corrected  according 
to  the  table  on  page  149.  The  given  m'iddle  latitude  is  to  be 
looked  for  either  in  the  first  or  last  vertical  column,  opposite 
to  which,  and  under  the  given  difference  of  latitude,  is  inserted 
the  proper  correction  in  minutes,  which  must  be  added  to  the 
middle  latitude  to  obtain  the  latitude  in  which  the  meridian 
distance  is  exactly  equal  to  the  departure.  Thus,  if  the  mid- 
dle latitude  is  41°,  and  the  difference  of  latitude  14°,  the  cor- 
rection will  be  found  to  be  25',  which,  added  to  the  middle 
latitude,  gives  the  corrected  middle  latitude  41°  25'. 


148  Trigonometry. 

Examples. 

1.  Find  tlio  bearing  and  distance  of  Liverpool,  latifade  52° 
22'  N.,  longitude  2°  52'  W.,  from  New  York,  latitude  40°  42'  N., 
longitude  74°  1'  W. 

Here  are  given  two  latitudes  and  longitudes  to  find  tha 
course  and  distance. 

The  difference  of  latitude  is      ....  12°  40'=  760'. 

The  difference  of  longitude  is   .     .     .     .  71°    9'=4269'. 

The  middle  latitude  is 47°    2'. 

To  which  add  the  correction  from  p.  149  22'. 

The  corrected  middle  latitude  is    .     .     .  47°  24'. 

Then,  according  to  the  third  of  the  preceding  theorems, 
Diff.  lat. :  cos.  mid.  lat. : :  diff.  long. :  tang".  cour<:e='N.  75°  16'  E 

To  find  the  distance  by  plane  sailing, 

Cos.  course  :  diff.  latitude  : :  R  :  distance— 298SA  miles. 

2.  A  ship  sailed  from  Bermuda,  latitude  32°  22'  N.,  longi- 
tude 64°  38'  W.,  a  distance  of  500  miles,  upon  a  course  W.N  W. 
Required  her  latitude  and  longitude  at  that  time. 

By  plane  sailing, 

R  :  distance  : :  cos.  course  :  diff.  latitude-  191.3. 

Therefore  the  required  latitude  is 35°  33' ; 

the  middle  latitude 33°  58' ; 

and  the  corrected  middle  latitude 33°  59' 

Then  we  have 

Cos.  mid.  lat.  :  distance  : :  sin.  course  :  diff.  long.=557'.l. 
Therefore  the  longitude  required  is  73°  55'. 

3.  A  ship  sails  southeasterly  from  Sandy  Hook,  latitude 
40°  28'  N.,  longitude  74°  1'  W.,  a  distance  of  395  miles,  when 
her  latitude  is  34°  40'-  N.     Required  her  course  and  longitude, 

Ans.  Course    S.  28°  14'  E. 
Longitude  70°    5' W. 

4.  A  ship  sails  from  Brest,  latitude  48°  23'  N.,  longitude 
4°  29'  W.,  upon  a  course  W.S.W.,  till  her  departure  is  556 
miles.     Required  the  distance  sailed  and  the  place  of  the  ship. 

Ans.  Distance  601.8  miles. 
Latitude  44°  33'  N. 
Longitude  17°  57'  W. 


Navigation. 


149 


MERCATOR'S  SAILING 


(201.)  Mercator's  sailing  is  a  method  of  computing  differ- 
ence of  bngitude  on  the  principles  of  Mercator's  chart.  On 
this  chart,  the  meridians,  instead  of  converging  toward  the 
poles  as  they  do  on  the  globe,  are  drawn  parallel  to  each  other, 
by  which  means  the  distance  of  the  meridians  is  every  wher<» 


150°       12.0        go"        60"        30°         0°         s 

0'       60°       go"     120      iSo" 

yi\.Mi__{% 

..  -  JL 

p       A 

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\^-^~KiT 

^  "^Xd: 

\  ^l     ^ 

J^     ^l^° 

■"  "'     '       -t       u 

it      r\ 

"E         ^ 

t . 

fs«    0                  ^ 

£>0    oO 

-■'       \_ 

6(f 

45  • 
30" 
15* 
0' 
15* 

30* 

do* 

15' 


made  too  great  except  at  the  equator.  To  compensate  for 
this,  in  order  that  the  outline  of  countries  may  not  be  too 
much  distorted,  the  degrees  of  latitude  are  proportionally  en- 
larged, so  that  the  distance  between  the  parallels  of  latitude 
increases  from  the  equator  to  the  poles.  In  latitude  60°  the 
distance  of  the  meridians  is  twice  as  great,  compared  with  a 
degree  at  the  equator,  as  it  is  upon  a  globe,  and  a  degree  of 
latitude  is  here  represented  twice  as  great  as  near  the  equator 
The  diameter  of  an  island  in  latitude  60°  is  represented  twice 
as  great  as  if  it  was  on  the  eg^uator,  and  its  area  four  times 
too  great.  In  latitude  70°  32'  the  distance  of  the  meridians 
is  three  times  too  great,  in  latitude  75°  31'  four  times  too  great 
and  so  on,  by  which  means  the  relative  dimensions  of  coun- 
tries in  high  latitudes  is  exceedingly  distorted.  On  this  ac- 
count it  is  not  common  to  extend  the  chart  beyond  latitude  75°. 
(202.)  The  distance  of  any  parallel  upon  Mercator's  chart 
from  the  equator  has  been  computed,  and  is  exhibited  in  the 


150  Trigonometry. 

accompanying  tables,  pages  142-8,  which  is  called  a  Table  of 
Meridional  Parts.  This  table  may  be  computed  in  the  fol- 
lowing  manner  : 

According  to  Art.  197,  cosine  of  latitude  is  to  radius,  as  the 
departure  is  to  the  difference  of  longitude ;  that  is,  as  a  part 
of  a  parallel  of  latitude  is  to  a  like  part  of  the  equator,  or  any 
meridian. 

But  by  Art.  28,  cosine  :  R  :  :  R  :  secant ;  hence 

V  of  a  parallel  :  V  of  a  meridian  :  :  R  :  sec.  latitude. 
But  on  Mercator's  chart  the  distance  between  the  meridiana 
is  the  same  in  all  latitudes  ;  that  is,  a  minute  on  a  parallel  of 
latitude  is  equal  to  a  minute  at  the  equator,  or  a  geographical 
mile.     Hence  the  length  of  one  minute,  on  any  part  of  a  me- 
ridian, is  equal  to  the  secant  of  the  latitude.     Thus, 
The  first  minute  of  the  meridian  =  the  secant  of  1' ; 
.  second         "•  «         =  "  2' , 

third  "  "         =  "  3', 

&c.,  &c. 

The  table  of  meridional  parts  is  "formed  by  adding  together 
the  minutes  thus  found.     Thus, 
Mer.  parts  of  l'=sec.  V  ; 
Mer.  parts  of  2'= sec.  I'+sec.  2' ; 
Mer.  parts  of  3'= sec.  I'+sec.  2'+ sec.  3' ; 
Mer.  parts  of  4'=sec.  I'+sec.  2'+sec.  3'+sec.  4', 
&c.,  &c.,  &c. 

Since  the  secants  of  small  arcs  are  nearly  equal  to  radius 
or  unity,  if  the  meridional  parts  are  only  given  to  one  tenth 
of  a  mile,  we  shall  have 

The  meridional  parts  of  l'=1.0  mile  ; 

a  a  a  2'  =  2.0      " 

"  a  u         3'  =  3.0     " 

"  "  "         4'=4.0     "     &o., 

as  shown  in  the  table  on  page  142. 

At  2°  33'  the  sum  of  the  small  fractions  omitted  becomes 
greater  than  half  of  one  tenth,  and  the  meridional  parts  of 
2°  33'  is  153.1 ;  that  is,  the  meridional  parts  exceed  by  one 
tenth  of  a  mile  the  minutes  of  latitude.  At  3°  40'  the  excess 
is  two  tenths  of  a  mile ;  at  4°  21'  the  excess  is  three  tenths ; 


Navigation.  151 

and  as  the  latitude  increases,  the  meridional  parts  increase 
more  rapidly,  as  is  seen  from  the  table. 

An  arc  of  Mercator's  meridian  contained  between  two  par- 
allels of  latitude  is  called  meridional  difference  of  latitude  Tt 
is  found  by  subtracting  the  meridional  parts  of  the  less  latitude 
from  the  meridional  parts  of  the  greater,  if  both  are  north  or 
south,  or  by  adding  them  together  if  one  is  north  and  the  other 
south.     Thus, 

The  lat.  of  New  York  is  40°  42' ;  meridional  parts=2677.8, 
"         New  Orleans  29°  57' ;  "  "        1884.9. 

The  true  diff.  of  lat.  is  10°  45' ;  mer.  difF.  lat.  is       792.9. 

If  one  latitude  and  the  meridional  difference  of  latitude  be 
given,  the  true  difference  of  latitude  may  be  found  by  reversing 
this  process.     Thus, 

The  meridional  parts  for  New  Orleans     .     .     .     =1884.9. 

Meridional  difference  of  latitude  between  New  )  _  ^qo  q 
York  and  New  Orleans 


Therefore  the  meridional  parts  for  New  York         =2677.8, 
and  the  corresponding  latitude  from  the  table  is  40°  42'. 

(203.)  If  we  take  the  figure  ABC  for 

plane  sailmg,  as  on  page  139,  and  pro-    ^| ^ 

duce  AC  to  E,  making  AE  equal  to  the 
meridional  difference  of  latitude,  then 
will  DE  represent  the  difference  of  lon- 
gitude corresponding  to  the  departure 
BC.  For  we  have  seen  (Art.  202)  that 
the  departure  is  to  the  difference  of  lon- 
gitude as  radius  is  to  the  secant  of  lati- 
tude, which  is  also  the  ratio  of  the  true 
difference  of  latitude  to  the  meridional  difference  of  latitude. 

Now,  from  the  similarity  of  the  triangles  ABC,  ADE,  we 
have 

AC  :  AE  : :  BC  :  DE  ; 

that  is,  the  true  difference  of  latitude  is  to  the  meridional  dif* 
ference  of  latitude^  as  the  departure  is  to  the  difference  of 
longitude. 

Also,  in  the  triangle  ADE,  we  have 

R  :  tan.  A  : :  AE  :  DE ; 


152  Trigonometry. 

that  is,  radius  is  to  the  tangent  of  the  course^  as  the  merid- 
ional difference  of  latitude  is  to  the  difference  of  longitude. 

Examples. 

1.  Find  the  bearing  and  distance  from  Sandy  Hook,  latitude 
40°  28'  N.,  longitude  74°  V  W.,  to  Havre,  latitude  49°  29'  N., 
longitude  0°  6'  E. 

The  true  difference  of  latitude  is  9°  V=  541' ; 

meridional  difference  of  latitude  =767.1 ; 

difference  of  longitude  is  74°  7' =4447. 

Hence,  to  find  the  course  by  the  preceding  proportion, 
Mer.  diff.  lat.  :  diff.  long.  :  :  R  :  tan.  course='N.  80°  13'  E. 
To  find  the  distance  by  plane  sailing, 
Cos.  course  :  true  diff.  lat.  :  :  R  :  distance=SlSS.S  miles. 

2.  Find  the  bearing  and  distance  from  Nantucket  Shoals,  in 
latitude  41°  4'  N.,  longitude  69°  55'  W.,  to  Cape  Clear,  in  lati- 
tude 51°  26'  N.,  longitude  9°  29'  W. 

Ans.  Course  N.  76°  E. 

Distance  2572.9  miles. 

3.  A  ship  sails  from  Sandy  Hook  a  distance  of  600  miles 
upon  a  course  S.  by  E.     Required  the  place  of  the  ship. 

The  difference  of  latitude  may  be  found  by  plane  sailing, 
the  difference  of  longitude  by  Mercator's  sailing. 

Ans.  Latitude    30°  39'.5  N. 
Longitude  71°  36'.7  W. 

4.  A  ship  sails  from  St.  Augustine,  latitude  29°  52'  N.,  lon- 
gitude 81°  25'  W.,  upon  a  course  N.E.  by  E.,  until  her  lati- 
tude is  found  to  be  34°  40'  N.  What  is  then  her  longitude, 
and  what  distance  has  she  run  ? 

Ans.  Longitude =72°  55'  W 
Distance    =518.4  miles. 

5.  A  ship  sails  from  Bermuda  upon  a  course  N.W.  by  W 
until  her  longitude  is  found  to  be  69°  30'  W.  What  is  then 
her  latitude,  and  what  distance  has  she  run  ? 

Ans.  Latitude  35°  4'  N. 
Distance  291.6  miles. 

6.  A  ship  sailing  from  Madeira,  latitude  32°  38'  N ,  longi- 
tude 16°  55'  W.,  steers  westerly  until  her  latitude  is  40°  2'  N., 


Natigation.  153 

and  her  departure  2425  miles.     Required  her  course,  distance 
tind  longitude. 

Ans.  Course  N.  79°  37'  W 
Distance  2465.3  miles. 
/   .  Longitude  67°  9'.3  W. 

7.  Find  the  bearing  and  distance  from  Sandy  Hook,  latitude 
40°  28'  N.,  longitude  74°  V  W.,  to  the  Cape  of  Good  Hope 
latitude  34°  22'  S.,  longitude  18°  30'  E. 

Ans.  Course 
Distance 


CHARTS. 

(204.)  The  charts  commonly  used  in  navigation  are  plane 
charts,  or  Mercator^s  chart.  In  the  construction  of  the  former, 
the  portion  of  the  earth's  surface  which  is  represented  is  sup- 
posed to  be  a  plane.  The  meridians  are  drawn  parallel  to  each 
other,  and  the  lines  of  latitude  at  equal  distances.  The  dis- 
tance between  the  parallels  should  be  to  the  distance  between 
the  meridians,  as  radius  to  the  cosine  of  the  middle  latitude 
of  the  chart.  A  chart  of  moderate  extent  constructed  in  this 
manner  will  be  tolerably  correct.  The  distance  of  the  merid- 
ians in  the  middle  of  the  chart  will  be  exact,  but  on  each  side 
it  will  be  either  too  great  or  too  small. 

When  large  portions  of  the  earth's  surface  are  to  be  repre- 
sented, the  error  of  the  plane  chart  becomes  excessive.  To 
obviate  this  inconvenience  Mercator's  chart  has  been  con- 
structed. Upon  this  chart  the  meridians  are  represented  by 
parallel  lines,  and  the  distance  between  the  parallels  of  latitude 
is  proportioned  to  the  meridional  difference  of  latitude,  as  rep- 
resented on  page  149. 

We  have  seen  that  the  meridional  difference  of  latitude  is  to 
the  difference  of  longitude  as  radius  is  to  the  tangent  of  the 
course.  Hence,  while  the  course  rem.ains  unchanged,  the  ratio 
of  the  meridional  difference  of  latitude  to  the  difference  of  lon- 
gitude is  constant ;  and,  therefore,  every  rhumb  line  will  be 
represented  on  Mercator's  chart  by  a  straight  line.  This 
property  renders  Mercator's  chart  peculiarly  ccnvenient  to 
navigators 


154  Trigonometry. 

The  preceding  sketch  affords  a  very  incomplete  view  of  the 
present  state  of  the  science  of  navigation.  The  most  accurate 
method  of  ascertaining  the  situation  of  a  vessel  at  sea  is  by 
means  cf  astronomical  observations.  For  these,  however,  the 
otud'jnt  must  be  referred  to  some  treatise  on  Astronomy. 


BOOK  VI. 

SPHERICAL  TRIGONOMETRY. 

(205.)  Spherical  trigonometry  teaches  how  to  determine 
the  several  parts  of  a  spherical  triangle  from  having  certain 
parts  given. 

A  spherical  triangle  is  a  portion  of  the  surface  of  a  sphere, 
bounded  by  three  arcs  of  great  circles,  each  of  which  is  less 
.than  a  semioircumference.  •  • 

RIGHT-ANGLED  SPHERICAL  TRIANGLES. 
JCheorem  I. 

(206.)  In  any  right-angled  spherical  triangle^  the  sine  of 
the  hypothenuse  is  to  radius^  as  the  sine  of  either  side  is  to 
the  sine  of  the  opposite  angle. 

Let  ABC  be  ^  spherical  triangle,  right-angled  at  A;  then 
will  the  sine  of  the  hypothenuse  BC 
be  to  radius,  as  the  sine  of  the  side 
AC  is  to  the  sine  of  the  angle  ABC. 

Let  D  be  the  center  of  the  sphere ; 
join  AD,  BD,  CD,  and  draw  CE  per- 
pendicular to  DB,  which  will,  there- 
fore, be  the  sine  of  the  hypothenuse 
BC.  From  the  point  E  draw  the 
straight  line  EF,  in  the  plane  ABD,  perpendicular  to  BD,  and 
join  CF.  Then,  because  DB  is  perpendicular  to  the  two  lines 
CE,  EF,  it  is  perpendicular  to  the  plane  CEF ;  and,  conse- 
quently, the  plane  CEF  is  perpendicular  to  the  plane  ABD 
(Geom.,  Prop.  6,  B.  YIL).  But  the  plane  CAD  is  also  per- 
pendicular to  the  plane  ABD  ;  therefore  their  line  of  common 
section,  CF,  is  perpendicular  to  the  plane  ABD ;  hence  CFD, 
CFE  are  right  angles,  and  CF  is  the  sine  of  the  arc  AC. 

Now,  in  the  right-angled  plane  triangle  CFE, 
CE  :  radius  :  :  CF  :  sine  CEF 


156  Trigonometry. 

But  since  CE  and  FE  are  both  at  right  angles  to  DB,  the 
angle  CEF  is  equal  to  the  inclination  of  the  planes  CBD,  ABD ; 
that  is,  to  the  spherical  angle  ABC.     Therefore, 
sine  BC  :  E,  :  :  sine  AC  :  sine  ABC. 

(207.)  Cor.  1.  In  any  right-angled  spherical  triangle,  t^p 
sines  of  the  sides  are  as  the  sines  of  the  opposite  angles. 

For,  by  the  preceding  theorem, 

sine  BC  :  R  :  :  sine  AC  :  sine  ABC, 
and  sine  BC  :  R  :  :  sine  AB  :  sine  ACB ; 

therefore,  sine  AC  :  sine  AB  :  :  sine  ABC  :  sine  ACB. 

Cor.  2.  In  any  right-angled  spherical  triangle,  the  cosine  of 
either  of  the  sides  is  to  radius,  as  the  cosine  of  the  hypothenuse 
is  to  the  cosine  of  the  other  side. 

Let  ABC  be  a  spherical  triangle,  right-angled  at  A.     De- 
scribe the  circle  DE,  of  which  B  is  p 
the  pole,  and  let  it  meet  the  three  \\ 
sides  of  the  triangle  ABC  produced  m  \   \ 
D,E,andF.     Then,  because  BD  and  \     \? 
BE  are  quadrants,  the  arc  DF  is  per-  I  /    \ 
pendicular  to  BD.     And  since  BAC                       ^''        \ 
is  a  right  angle,  the  arc  AF  is  per- 
pendicular to  BD.     Hence  the  point 
F,  where  the  arcs  FD,  FA  intersect                       -^ 
each  other,  is  the  pole  of  the  arc  BD  (Geom.,  Prop.  5,  Cor.  2, 
B.  IX.),  and  the  arcs  FA,  FD  are  quadrants. 

Now,  in  the  triangle  CEF,  right-angled  at  the  point  E,  ac- 
cording to  the  preceding  theorem,  we  have 

sine  CF  :  R  :  :  sine  CE  :  sine  CFE. 

But  CF  is  the  complement  of  AC,  CE  is  the  complement  of 
BC,  and  the  angle  CFE  is  measured  by  the  arc  AD,  which  is 
the  complement  of  AB.  Therefore,  in  the  triangle  ABC,  we 
have 

cos.  AC  :  R  : :  cos.  BC  :  cos.  AB. 

Cor.  3.  In  any  right-angled  spherical  triangle,  the  cosine 
of  either  of  the  sides  is  to  radius,  as  the  cosine  of  the  angle 
opposite  to  that  side  is  to  the  sine  of  the  other  angle. 

For,  in  the  triangle  CEF,  we  have 

sine  CF  :  R  :  :  sine  EF  :  sine  ECF. 

But  sine  CF  is  equal  to  cos.  CA.     EF  is  the  complement  of 


■<:i> 


m 

Spherical   Trigonometry.  157 

ED,  which  measures  the  angle  ABC,  that  is,  sine  EF  is  equal 
to  COS.  ABC,  and  sine  ECF  is  the  same  as  sine  ACB ;  there- 
fore, 

COS.  AC  :  R  :  :  cos.  ABC  :  sine  ACB 

Theorem  II. 

(208.)  In  any  right-angled  spherical  triangle^  the  sine  oj 
either  of  the  sides  about  the  right  angle  is  to  the  cotangent 
of  the  adjacent  angle,  as  the  tangent  of  the  remaining  side 
is  to  radius. 

Let  ABC  be  a  spherical  triangle,  right-angled  at  A ;  then 
will  the  sine  of  the  side  AB  be  to  the  -ni 

cotangent  of  the  angle  ABC,  as*  the  y 

tangent  of  the  side  AC  is  to  radius.  Cy^ 

Let  D  be  the  center  of  the  sphere ;  y^   \H 

join  AD,  BD,  CD  ;  draw  AE  perpen-  y^         Y 

dicular  to  BD,  which  will,  therefore,  >/  /) 

be  the  sine  of  the  arc  AB.    Also,  from  d-^™ /— t- 

the  point  E  in  the  plane  BDC,  draw         ^^^^.^^^       ,4'"^/ 
the  straight  line  EF  perpendicular  to  ^^^^^ 

BD,  meeting  DC  produced  in  F,  and 

join  AF.  Then  will  AF  be  perpendicular  to  the  plane  ABD, 
because,  as  was  shown  in  the  preceding  theorem,  it  is  the  com- 
mon section  of  the  two  planes  ADF,  AEF,  each  perpendiculai 
to  the  plane  ADB.  Therefore  FAD,  FAE  are  right  angleai 
and  AF  is  the  tangent  of  the  arc  AC. 

Now,  in  tho^ triangle  AEF,  right-angled  at  A,  we  have 
AE  :  radius  :  :  AF  :  tang.  AEF. 

But  AE  is  the  sine  of  the  arc  AB,  AF  is  the  tangent  of  tL- 
arc  AC,  and  the  angle  AEF  is  equal  to  the  inclination  of  tH 
nlanes  CBD,  ABD,  or  to  the  spherical  angle  ABC ;  hence 
sine  AB  :  R  :  :  tang.  AC  :  tang.  ABC. 

And  because.  Art.  28, 

R  :  cot.  ABC  : :  tang.  ABC  :  R ; 
therefore,  sine  AB  :  cot.  ABC  : :  tang.  AC     :  R. 

(209.)  Cor.  1.  In  any  right-angled  spherical  triangle,  tkt 
cosine  of  the  hypothenuse  is  to  the  cotangent  of  either  of  the 
oblique  angles,  as  the  cotangent  of  the  other  oblique  ang-,t  is 
to  radius 


158  Trigonometry. 

Let  ABC  be  a  spherical  triangle,  right-angled  at  A,     De- 
scribe the  circle  DEF,  of  which  B  p 
is  the  pole,  and  construct  the  com-  \\^ 
plemental  triangle  CEF,  as  in  Cor.  \  \ 
2,  Theorem  I.  \     \e 

Then,  in  the  triangle  CEF,  ac-  \  /  \ 

3ording  to  the  preceding  theorem,  we  yf        \ 

sine  CE  :  cot.  ECF  : :  tan.  EF  :  R.        ^^-^___Z..---"' 
But  CE  is  the  complement  of  BC,         ^  A 

EF  is  the  complement  of  ED,  the  measure  of  the  angle  ABC ; 
and  the  angle  ECF  is  equal  to  ACB,  being  its  vertical  angle; 
hence 

COS.  BC  :  cot.  ACB  : :  cot.  ABC  :  R. 
Cor.  2.  In  any  right-angled  spl^rical  triangle,  the  cosine  of 
either  of  the  oblique  angles  is  to  the  tangent  of  the  adjacent 
side,  as  the  cotangent  of  the  hypothenuse  is  to  radius. 

For,  in  the  complemental  triangle  CEF,  according  to  the 
preceding  theorem,  we  have 

sine  EF  :  cot.  CFE  :  :  tan.  CE  :  R ; 
hence,  in  the  triangle  ABC, 

COS.  ABC  :  tan.  AB  :  :  cot.  BC  :  R. 

Napier'' s  Rule  of  the  Circular  Parts. 

(210.)  The  two  preceding  theorems,  with  their  corollaries, 
"are  sufficient  for  the  solution  of  all  cases  of  right-angled  spheri- 
cal triangles,  and  a  rule  was  invented  by  Napier  by  means  of 
which  these  principles  are  easily  retained  in  mind. 

If,  in  a  right-angled  spherical  triangle,  we  set  aside  the  right 
angle,  and  consider  only  the  five  remaining  parts  of  the  trian- 
gle, viz.,  the  three  sides  and  the  two  oblique  angles,  then  the 
two  sides  which  contain  the  right  angle,  and  the  complements 
of  the  other  three,  viz.,  of  the  two  angles  and  the  hypothenuse, 
are  called  the  circular  parts. 

Thus,  in  the  triangle  ABC,  right-angled  at  A,  the  circular 
parts  are  AB,  AC,  with  the  complements  of  B,  BC,  and  C. 

"When,  of  the  five  circular  parts,  any  one  is  taken  for  the 
middle  part,  then,  of  the  remaining  four,  the  two  which  are 
immediately  adjacent  to  it  on  the  right  and  left  are  called  the 


Spherical   Trigonometry.  159 

adjaceni  parts  ;  and  the  other  two,  each  of  which  is  separated 
from  the  middle  by  an  adjacent  part, 
are  called  opposite  parts. 

In  every  question  proposed  for  solu- 
tion, three  of  the  circular  parts  are 
concerned,  two  of  which  are  given, 
and  one  required  ;  and  of  these  three, 
the  middle  part  must  be  such  that 
the  other  two  may  be  equidistant  from  it ;  that  is,  may  be 
either  both  adjacent  or  both  opposite  parts.  The  value  of  the 
part  required  may  then  be  found  by  the  following 

Rule  of  Napier. 
(211.)  The  product  of  the  radius  and  the  sine  of  the  middle 
part^  is  equal  to  the  product  of  the  tangents  of  the  Sidjacent 
parts,  or  to  the  product  of  the  cosines  of  the  opposite  parts. 

It  will  assist  the  learner  in  remembering  this  rule  to  remark, 
that  the  first  syllable  of  each  of  the  words  tangent  and  adja- 
cent contains  the  same  vowel  a,  and  the  first  syllable  of  the 
words  cosine  and  opposite  contains  the  same  vowel  o. 

It  is  obvious  that  the  co&ine  of  the  complement  of  an  angle 
is  the  sine  of  that  angle,  and  the  tangent  of  a  complement  la 
a  cotangent,  and  vice  versa. 

In  the  triangle  ABC,  if  we  take  the  side  b  as  the  middle 
part,  then  the  side  c  and  the  complement  of  the  angle  C  are 
the  adjacent  parts,  and  the  complements  of  the  angle  B  and  of 
the  hypothenuse  a  are  the  opposite  parts.  Then,  according  to 
Napier's  rule,  R  sin.  6=tan.  c  cot.  C, 

which  corresponds  with  Theorem  II. 
Also,  by  Napier's  rule,- 

R  sin.  6= sin.  a  sin.  B, 
which  corresponds  with  Theorem  I. 

Making  each  of  the  circular  parts  in  succession  the  middle 
part,  we  obtain  the  ten  following  equations : 

R  sin.  6=sin.  a  sin.  B=tan.  c  cot.  C. 
R  sin.  c=sin.  a  sin.  C=tan.  b  cot.  B. 
R  COS.  B=cos.  b  sin.  C=cot.  a  tan.  c. 
R  COS.  a=GOB,  b  COS.  c  =oot.  B  cot.  C. 
R  COS.  C=cos.  c  sin.  B=cot.  a  tan.  b. 


160  Trigonometry. 

(212.)  In  order  to  determine  whether  the  quantity  sought 
is  less  or  greater  than  90°,  the  algebraic  sign  of  each  term 
should  be  preserved  whenever  one  of  them  is  negative.  If  the 
quantity  sought  is  determined  by  means  of  its  cosine,  tangent, 
or  cotangent,  the  algebraic  sign  of  the  result  will  show  w^hether 
this  quantity  is  less  or  greater  than  90° ;  for  the  cosines,  tan- 
gents, and  cotangents  are  positive  in  the  first  quadrant,  and 
negative  in  the  second.  But  since  the  sines  are  positive  in  both 
the  first  and  second  quadrants,  when  a  quantity  is  determined 
by  means  of  its  sine,  this  rule  will  leave  it  ambiguous  whether 
the  quantity  is  less  or  greater  than  90°.  The  ambiguity  may, 
however,  generally  be  removed  by  the  following  rule. 

In  every  right-angled  spherical  triangle,  an  oblique  angle 
and  its  opposite  side  are  always  of  the  same  species ;  that  is, 
both  are  greater,  or  both  less  than  90°. 

This  follows  from  the  equation 

R  sin.  6= tan.  c  cot.  C ; 
where,  since  sin.  h  is  always  positive,  tan.  c  must  always  have 
the  same  sign  as  cot.  C ;  that  is,  the  side  c  and  the  opposite 
angle  C  both  belong  to  the  same  quadrant. 

(218.)  When  the  given  parts  are  a  side  and  its  opposite  an- 
gle, the  problem  admits  of  two  solutions  ;  for  two  right-angled 
spherical  triangles  may  always  be  found,  having  a  side  and  its 
opposite  angle  the  same  in  both,  but  of  which  the  remaining 
sides  and  the  remaining  angle  of  the  one  are  the  supplements 
of  the  remaining  sides  and  the  remaining  angle  of  the  other. 
Thus,  let  BCD,  BAD  be  the  halves  of  two  great  circles,  and 
let  the  arc  CA  be  drawn  perpendicu- 
lar to  BD  ;  then  ABC,  ADC  are  two 
right-angled  triangles,  having  the  side  B  - 
AC  common,  and  the  opposite  angle 
B  equal  to  the  angle  D ;  but  the  side  DC  is  the  supplement  of 
BC,  AD  is  the  supplement  of  AB,  and  the  angle  ACD  is  the 
supplement  of  ACB. 

Examples. 
1.  In  the  right-angled  spherical  triangle  ABC,  there  are 
given  a =63°  56'  and  6=40°.     Required  the  other  side  c,  and 
the  angles  B  and  C 


Spherical    Trigonometry.  161 

'It  tind  the  side  c. 

Hei^N'  the  circular  parts  concerned  are 
the  two  legs  and  the  complement  of  the 
hypothenase;  and  it  is  evident  that  if 
the  compL:5ment  of  a  be  made  the  mid- 
dle part^  b  and  c  will  be  opposite  parts  ; 
hence,  by  Napier's  rule, 

E,  COS.  a=^cos.  b  cos.  c ; 
or,  reducing  this  equation  to  a  proportion, 

COS.  ^  :  R  :  :  cos.  a  :  cos.  c=54°  59'  49". 

To  find  the  angle  B. 

Here  b  is  the  middle  part^  and  the  complements  of  B  and  a 
are  opposite  parts  ;  hence 

R  sin.  ^=cos.  (comp.  a)Xcos.  (comp.  B)=sin.  a  sin.  B, 
or  sin.  a  :  R  :  :  sin.  b  :  sin.  B=45°  41'  25". 

B  is  known  to  be  an  acute  angle,  because  Us  opposite  side  is 
less  than  90°. 

To  find  the  angle  C. 

Here  the  complement  of  C  is  the  middle  part;  also  b  and 
the  complement  of  a  are  adjacent  parts  ;  hence 

R  COS.  C=cot.  a  tan.  b, 
or  R  :  tan.  b  :  :  cot.  a  :  cos.  C=65°  45'  57". 

Ex.  2.  In  a  right-angled  triangle  ABC,  there  are  given  the 
hypothenuse  a =91°  42',  and  the  angle  B=95°  6'.  Required 
the  remaining  parts. 

To  find  the  angle  C. 

Make  the  complement  of  the  hypothenuse  the  middle  part; 
then  R  cos.  a=cot.  B  cot.  C. 

Whence  C=71°  36'  47". 

To  find  the  side  a. 

Make  the  complement  of  the  angle  B  the  middle  part ;  and 
we  have  R  cos.  B=cot.  a  tan.  c. 

Whence  c=71°  32'  14 '. 

To  find  the  side  b. 

Make  the  side  b  the  middle  part ;  then 

L 


162  Trigonometry. 

R  sin.  ^=sin.  a  sin.  B. 
Whence  -  ^>=95°  22'  30". 

b  is  known  to  be  greater  than  a  quadrant,  because  its  opposite 
angle  is  obtuse. 

Ex.  3.  In  the  right-angled  triangle  ABC,  the  side  6  is  26"* 
4',  and  its  opposite  angle  B  36°.  Required  the  remaining 
parts. 

'      i  a  =48°  22'  52",  or  131°  37'    8"' 
Ans.  \  c  =42°  19'  17",  or  137°  40'  43". 
(  C=64°  14'  26",  or  115°  45'  34". 
This  example,  it  will  be  seen,  admits  of  two  solutions,  con- 
formably to  Art.  213. 

Ex.  4.  In  the  right-angled  spherical  triangle  ABC,  there  are 
given  the  side  c,  54°  30',  and  its  adjacent  angle  B,  44°  50'. 
Required  the  remaining  parts. 

(  C=65°  49'  53'^ 
il;^5.  )a=63°  10'    4". 
(  b  =38°  59'  11" 
Why  is  not  the  result  ambiguous  in  this  case  ? 
Ex.  5.  In  the  right-angled  spherical  triangle  ABC,  the  side 
b  is  6^°  28',  and  the  side  c  63°  15'.     Required  the  remaining 
parts.  • 

..'  /yr--  r  a=75°  13'    2". 

[\(v/y'f  _  Ans.  \  B=58°  25'  47" 

vAvlil'^^  (C=67°27'    1". 

Ex.  6.  In  the  right-angled  spherical  triangle  ABC,  there  are 
given  the  angle  B=69°  20',  and  the  angle  C=58°  16'.  Re- 
quired the  remaining  parts. 

(  a=76°  30'  37". 
Ans.  ]^>=65°  28' 58". 
(  c  =65"  47'  46". 
(214.)  A  triangle,  in  which  one  of  the  sides  is  equal  to  a 
quadrant,  may  be  solved  upon  the  same  principles  as  right- 
angled  triangles,  for  its  polar  triangle  will  contain  a  right  an- 
gle.    See  Geom.,  Prop.  9,  B.  IX. 

Ex.  7.  In  the  spherical  triangle  ABC,  the  side  BC=90°,  the 
angle  C=42°  10',  and  the  angle  A=115°  20'.  Required  the 
remaining  parts. 

Taking  the  supplements  of  the  given  parts,  we  shall  have 


Spherical   Trigonometry 


163 


in  the  polar  triangle  the  hypothenuse  a' =180''— 115°  20' =64'' 
40',  and  one  of  the  sides,  c'=180°-42°  10'=137°  50',  from 
which,  by  Napier's  rule,  we  find 

B'=115°  23'  20". 
0=132°    2'  13". 
b'=125^  15' 36". 
Hence,  taking  the  supplements  of  these  arcs,  we  find  the 
parts  of  the  required  triangle  are 

AC  =  64°  36'  40". 

AB=47°  57'  47". 

B    =54°  44'  24". 

Ex.  8.  In  the  spherical  triangle  ABC,  the  side  AC=90°,  the 

angle  C=69°  13'  46",  and  the  angle  A=72°  12'  4".     Required 

the  remaining  parts. 

rAB=70°    8'39'^ 
^W5.  ]BC=73°  17' 29". 
(  B    =96°  13'  23 


OBLIQUE-ANGLED  SPHERICAL  TRIANGLES. 
Theorem  III. 
(215.)  In  any  spherical  triangle.^  the  sines  of  the  sides  are 
proportional  to  the  sines  of  the  opposite  angles. 

In  the  case  of  right-angled  spherical  triangles,  this  proposi- 
tion has  already  been  demonstrated. 
Let,  then,  ABC  be  an  oblique-angled 
triangle  ;  we  are  to  prove  that 
sin.  BC  :  sin.  AC  : :  sin.  A  :  sin.  B. 
Through  the  point  C  draw  an  arc 
of  a  great  circle  CD  perpendicular  to 
AB.     Then,  in  the  spherical  triangle 
ACD,  right-angled  at  D,  we  have,  by  Napier's  rule, 
R  sin.  CD = sin.  AC  sin.  A. 
Also,  in  the  triangle  BCD,  we  have 

R  sin.  CD=sin.  BC  sin.  B. 
Hence         sin  AC  sin.  A=sin.  BC  sin.  B, 
or  sin.  BC  :  sin.  AC  : :  sin.  A  :  sin.  B. 

(216.)  Cor.  1.  In  any  spherical  triangle,  the  cosines  of  the 
sides  are  proportional  to  the  cosines  of  the  segments  of  the 
base,  made  by  a  perpendicular  from  the  opposite  angle. 


164 


Trigonometry. 


For,  by  Theorem  I.,  Cor.  2, 

COS.  CD  :  R  :  :  cos.  AC  :  cos.  AD. 
Also,  COS.  CD  :  R  :  :  cos.  BC  :  cos.  BD. 

Hence  cos.  AC  :  cos.  BC  :  :  cos.  AD  :  cos.  BD. 

Cor.  2.   The  cosines  of  the  angles  at  the  base  are  propor- 
tional to  the  sines  of  the  segments  of  the  vertical  angle. 
For,  by  Theorem  I.,  Cor.  3, 

COS.  CD  :  R  :  :  cos.  A  :  sin.  ACD. 
Also,  COS.  CD  :  R  :  :  cos.  B  :  sin.  BCD. 

Hence         cos.  A :  cos.  B  : :  sin.  ACD  :  sin.  BCD. 

Cor.  3.   The  sines  of  the  segments  of  the  base  are  recipro- 
cally proportional  to  the  tangents  of  the  angles  at  the  basp 

For,  by  Theorem  H., 

sin.  AD  :  R  : :  tan.  CD  :  tan.  A. 

Also,  sin.  BD  :  R  :  :  tan.  CD  :  tan.  B. 

Hence  sin.  AD  :  sin.  BD  :  :  tan.  B  :  tan.  A. 

Cor.  4.  The  cotangents  of  the  two  sides  are  proportional 
to  the  cosines  of  the  segments  of  the  vertical  angle. 

For,  by  Theorem  H.,  Cor.  2, 

cos.  ACD  :  cot.  AC  :  :  tan.  CD  :  R. 

Also,         COS.  BCD  :  cot.  BC  :  :  tan.  CD  :  R. 

Hence       cos.  ACD  :  cos.  BCD  :  :  cot.  AC  :  cot.  BC. 


Theorem  IV. 

(217.)  If  from  an  angle  of  a  spherical  triangle  a  perpen^ 
dicular  be  drawn  to  the  base,  then  the  tangent  of  half  the  sum 
of  the  segments  of  the  base  is  to  the  tangent  of  half  the  sum 
of  the  sides,  as  the  tangent  of  half  the  difference^  of  tlie  sides 
is  to  the  tangent  of  half  the  difference  of  the  segments  of  the 
base. 

Let  ABC  be  any  spherical  trian- 
gle, and  let  CD  be  drawn  from  C 
perpendicular  to  the  base  AB  ;  then 
tan.  i(BD+AD) :  tan.  i(BC+AC) : : 
tan.  i(BC-AC)  :  tan.  i(BD-AD). 

Let  BC=«,  AC=^  BD=m,  and  ^' 
AD=^w.     Then,  by  Theorem  UL,  Cor.  1, 

cos.  a  :  cos.  b  :  :  cos.  m  :  cos. 


Spherical   Trigonometry.  .      166 

WhBiice,  Greom.,  Prop.  7,  Cor.,  B.  II., 
COS.  b+cos.  a  :  cos.  Z>— cos.  a  : :  cos.  »+cos.  m  :  cos.  n—cos.  m 

But  by  Trig.,  Art.  76, 
cos.  b+cos.  a  :  cos.  6— cos.  a  : :  cot.  \{a-\-b)  :  tan.  \(a'-b). 

Also,  by  the  same  Art., 
cos.  w+cos.  m  :  cos.  7Z— cos.  m  :  :  cot.  \{m-\-n)  :  tan.  J(m— »). 

Therefore 

cot.  \(a-]rb)  :  cot.  ^(m+n)  :  :  tan.  i(«— ^)  :  tan.  ^m—n). 

But,  since  tangents  are  reciprocally  as  their  cotangents.  Art. 
28,  we  have 

cot.  ^(a+b)  :  cot.  ^m+n)  :  :  tan.  ^(m+n)  :  tan.  ^(a+b) 

Hence 

tan.  ^(m+n)  :  tan.  ^{a+b)  :  :  tan.  l(a—b)  :  tan.  ^(m— w). 

(218.)  In  the  solution  of  oblique-angled  spherical  triangles, 
six  cases  may  occur,  viz. : 

1.  Griven  two  sides  and  an  angle  opposite  one  of  them. 

2.  Griven  two  angles  and  a  side  opposite  one  of  them. 

3.  Griven  two  sides  and  the  included  angle. 

4.  Griven  two  angles  and  the  included  side. 

5.  Given  the  three  sides. 

6.  Given  the  three  angles.    . 

Case  I. 

(219.)  Given  ttho  sides  and  an  angle  opposite  one  of  them, 
to  find  the  remaining  parts. 

In  the  triangle  ABC,  let  there  be  given  the  two  sides  AC 
and  BC,  and  .the  angle  A  opposite  one  ^ 

of  them.  •  The  angle  B  may  be  found 
by  Theorem  III. 
sin.  BC  :  sin.  AC  : :  sin.  A  :  sin.  B. 

From  the  angle  C  let  fall  the  per- 
pendicular CD  upon  the  side  AB. 
The  triangle  ABC  is  divided  into  two  right-angled  triangles, 
in  each  of  which  there  is  given  the  hypothenuse  and  the  angle 
at  the  base.  The  remaining  parts  may  then  be  found  by  Na- 
pier's rule.  • 

Ex.  1.  In  the  oblique-angled  spherical  triangle  ABC,  the 


166      ^  Trigonometry. 

Bide  AC=70°  10'  30",  BC=80°  5'  4",  and  the  angle  A=33^ 
15'  7".     Required  the  other  parts. 

sin  BC  :  sin.  AC  : :  sin.  A  :  sin.  B=31°  34'  38". 
Then,  in  the  triangle  ACD, 

R  COS.  AC = cot.  A  cot.  AQD. 
Whence  ACD=77°  27'  47". 

Also,  in  the  triangle  BCD, 

R  COS.  BC=cot.  B  cot.  BCD. 
Whence  BCD=  83°  57'  29". 

Therefore  ACB=161°  25'  16". 

To  find  the  side  AB. 

sin.  A  :  sin.  ACB  : :  sin.  BC  :  sin.  AB=145°  5'  0". 

When  we  have  given  two  sides  and  an  opposite  angle,  there 
are,  in  general,  two  solutions,  each  of  which  will  satisfy  the 
conditions  of  the  problem.  If  the  side  AC,  the  angle  A,  and 
the  side  opposite  this  angle  are  given, 
then,  with  the  latter  for  radius,  de- 
scribe an  arc  cutting  the  arc  AB  in 
the  points  B  and  B'.  The  arcs  CB, 
CB'  will  be  equal,  and  each  of  the  tri- 
angles ACB,  ACB'  will  satisfy  the 
conditions  of  the  problem.  There  is  the  same  ambiguity  in 
the  numerical  computation.  The  angle  B  is  found  by  moans 
of  its  sine.  But  this  may  be  the  sine  eithel*  of  ABC,  or  of  its 
supplement  AB'C  (Art.  27).  In  the  preceding  example,  the 
first  proportion  leaves  it  ambiguous  whether  the  angle  B  is 
31°  34'  38',  or  its  supplement  148°  25'  22".  In  oyder  to  avoid 
false  solutions,  we  should  remember  that  the  greater  side  of 
a  spherical  triangle  must  lie  opposite  the  greater  angle,  and 
conversely  (Geom.,  Prop.  17,  B.  IX.).  Thus,  since  in  the  pre- 
ceding example  the  side  AC  is  less  than  BC,  the  angle  B  must 
be  less  than  A,  and,  therefdi*e,  can  not  be  obtuse. 

If  the  quantity  sought  is  determined  by  means  of  its  cosine, 
tangent,  or  cotangent,  the  algebraic  sign  of  the  result  will 
show  whether  this  quantity  is  less  or  greater  than  90°  ;  for  the 
cosines,  tangents,  and  cotangents  are  positive  in  the  first  quad- 
rant, and  negative  in  the  second.     Hence  the  algebraic  sign 


Spherical   Trigoncmetry.  167 

f 

of  each  term  of  a  proportion  should  be  preserved  whenever  one 

of  them  is  negative. 

Ex.  2.  In  the  spherical  triangle  ABC,  the  side  a=124°  53', 

b=2V  19',  and  the  angle  A=16°  26'.     Required  the  remain- 

ing  parts. 

(  B=  10°  19'  34" 

Ans.l  C=171°  48' 22" 

(  c  =155°  35'  22" 

Case  II. 
(220.)  Given  two  angles  and  a  side  opposite  one  of  them^ 
to  find  the  remaining  parts. 

In  the  triangle  ABC  let  there  be  given  two  angles,  as  A  and 
B,  and  the  side  AC  opposite  to  one 
of  them.      The  side  BC   may  be  y\ 

found  by  Theorem  III.  //    \v 

sin.  B  :  sin.  A  : :  sin.  AC  :  sin.  BC.       /     j"  ^"^^ 

From  the  unknown  ansjle  C  draw        ^^--i.^^  _^.^^^ 

CD  perpendicular  to  AB  ;  then  will 

the  triangle  ABC  be  divided  into  two  right-angled  triangles,  in 
each  of  which  there  is  given  the  hypothenuse  and  the  angle  at 
the  base.  Whence  we  may  proceed  by  Napier's  rule,  as  in 
Case  I. 

Ex.  1.  In  the  oblique-angled  spherical  triangle  ABC,  there 
are  given  the  angle  A=52°  20',  B=63°  40',  and  the  side 
^=83°  25'.     Required  the  remaining  parts. 

sin.  B  :  sin.  A  :  :  sin.  AC  :  sin.  BC=61°  19'  53".       - 
Then,  in  the  triangle  ACD, 

cot.  IC  :  R  :  :  cos.  A  :  tan.  AD=79°  18'  17". 
Also,  in  the  triangle  BCD, 

cot.  BC  :  R  :  :  cos.  B  :  tan.  BD=39°  3'  8". 
Hence  AB=118°  21'  25". 

To  find  the  angle  ACB. 
sin.  BC  :  sin.  AB  : :  sin.  A  :  sin.  ACB=127°  26'  47". 

When  we  have  given  two  angles  and  an  opposite  side,  there 
are,  in  general,  two  solutions,  each  of  which  wiir satisfy  the 
conditions  of  the  problem.     If  the  angle  A,  the  side  AC,  and 


168  Trigonometry. 

the  angle  opposite  this  side  are  given,  then  through  the  point 
C  there  may  generally  be 
drawn  two  arcs  of  great  cir-  ^v 

oles  CB,  CB',  making  the  /       \:**^-.  B' 

same  angle  with  AB,  and        /  ^^^-^  /^ 

each  of  the  triangles  ABC,   ^  \.^^^^  ^^^^^^B 

AB'C  will  satisfy  the  condi- 
tions of  the  problem.  There  is  the  same  ambiguity  in  the 
numerical  computation,  since  the  side  BC  is  found  by  means 
of  its  sine  (Art.  27).  In  the  preceding  example,  however, 
there  is  no  ambiguity,  because  the  angle  A  is  less  than  B, 
and,  therefore,  the  side  a  must  be  less  than  b,  that  is,  less  than 
a  quadrant. 

Ex.  2.  In  the  oblique-angled  spherical  triangle  ABC,  the 
angle  A  is  128°  45',  the  angle  C=30°35^  and  BC  =  68o50'. 
Uequired  the  remaining  parts. 

It  will  be  observed  that  in  this  case  the  perpendicular  BD, 
drawn  from  the  angle  B,  falls  without  the  triangle  ABC,  and 
therefore  the  side  AC  is  the  difference  between  the  segments 
CD  and  AD.  (  AB=37°  28'  20' 

Ans,  \  AC =40°    9'    4". 
(B    =32°  37' 58". 

Case  III. 

(221.)  Given  two  sides  and  the  included  angle,  to  find  the 
remaining'  parts. 

In  the  triangle  ABC  let  there  be  given  two  sides,  as  AB, 
AC,  and  the  included  angle  A.     Let  ^ 

fall  the  perpendicular  CD  on  the  side 
AB  ;  then,  by  Napier's  rule, 

R  cos.  A=tan.  AD  cot.  AC. 

Having  found  the  segment  AD,  the 
segment  BD  becomes  known;  then, 
by  Theorem  III.,  Cor.  3, 

sin.  BD  :  sin.  AD  :  :  tan.  A  :  tan.  B. 

The  remaining  parts  may  now  be  found  by  Theorem  III. 

Ex.  1.  In  the  spherical  triangle  ABC,-  the  side  AB=73°  20  , 
AC=41°  45',  and  the  angle  A=30°  30'.  Required  the  remain- 
ing  parts. 


Spheric  A. L   Trigonometry.  169 

cot.  AC  :  COS.  A  :  :  R  :  tan.  AD=37°  33'  41". 
Hence  BD=35°  46'  19". 

sin.  BD  :  sin.  AD  : :  tan.  A.  :  tan.  B=31°  33'  43". 
Also,  by  Theorem  III.,  Cor.  1, 

cos.  AD  :  COS.  BD  :  :  cos.  AC  :  cos.  BC=40^  13^  0'\ 
Then,  by  Theorem  III., 

sin.  BC  :  sin.  AB  :  :  sin.  A  :  sin.  ACB=131°  8"  AT . 
Ex.  2.  In  the  spherical  triangle  ABC,  the  side  AB=78°  15', 
AC=56°  20',  and  the  angle  A=120°.     Required  the  other 
parts. 

r  B   =48°  57'  29". 

Ans.  ]g    =62°  31'  40". 

(BC=107°    7' 45" 

Case  IV. 
(222.)  Given  two  angles  and  the  included  side,  to  find  tne 
remaining-  parts. 

In  the  triangle  ABC  let  there  be  given  two  angles,  as  A  and 
ACB,  and  the  side  AC  included  be- 
tween them.     From  C  let  fall  the  per- 
pendicular CD  on  the  side  AB.     Then, 
by  Napier's  rule, 

R  COS.  AC = cot.  A  cot.  ACD. 
Having  found  the  angle  ACD,  the 
angle  BCD  becomes  known ;  then,  by 
Theorem  III.,  Cor.  4, 

COS.  ACD  :  COS.  BCD  :  :  cot.  AC  :  cot.  BC. 
The  remaining  parts  may  now  be  found  by  Theorem  III. 
Ex.  1.  In  the  spherical  triangle  ABC,  the  angle  A=32°  10', 
the  angle  ACB =133°  20',  and  the  side  AC =39°  15'.     Re- 
quired the  other  parts. 

cot.  A  :  cos.  AC  :  :  R  :  cot.  ACD=64°  1'  57". 
Hence  BCD=69°  18'  3". 

Then 

cos.  ACD  :  cos.  BCD  : :  cot.  AC  :  cot.  BC=45°  20'  43". 
Also,  by  Theorem  III.,  Cor.  2, 

sin.  ACD  :  sin.  BCD  : :  cos.  A  :  cos.  B=28°  15'  47". 
Then,  by  Theorem  III., 

sin.  B  :  sin  ACB  :  :  sin.  AC  :  sin.  AB=76°  23'  5". 


170  Trigonometry. 

Ex.  2.  In  the  ppherioal  triangle  ABO,  the  angle  A=125°  20', 
the  angle  0=48'^  30',  and  the  side  AC=83°  13'.  Reauired 
^he  remaining  parts. 

(  AB=  56°  39'    9". 

Ans.  )  BC  =  114°  30'  24". 

(  B    =62°  54'  38". 

Case  Y. 
(223.)  Given  the  three  sides  of  a  spherical  triangle,  to  find 
the  angles. 

In  the  triangle  ABC  let  there  be  given  the  three  sides. 
From  one  of  the  angles,  as  C,  draw  CD 
perpendicular  to  AB.     Then,  by  The- 
orem IV.,  tan.  |AB  :  tan.  |(AC+BC)  : 
tan.  i(AC-BC)  :  tan.  4(AD-BD). 

Hence  AD  and  BD  become  known ; 
then,  by  Napier's  rule, 

R  COS.  A=tan.  AD  cot.  AC. 
The  other  angles  may  now  be  easily  found. 
It  is  generally  most  convenient  to  let  fall  the  perpendiculaJ 
upon  the  longest  side  of  the  triangle. 

Ex.  1.  In  the  spherical  triangle  ABC,  the  side  AB=112*' 
25',  AC =60°  20',  and  BC=81°  10'.     Required  the  angles. 
tan.  6Q''  12J'  :  ^an.  70°  45'  :  :  tan.  10°  25'  :  tan.  19°  24'  26". 
Hence     AD=36°  48'  4",  and  BD=75°  36'  66". 
Then    R  :  tan.  AD  :  :  cot.  AC  :  cos.  A=64°  46'  36". 
Also,     R  :  tan.  BD  :  :  cot.  BC  :  cos.  B=52°  42'  12". 
Then  sin.  AC  :  sin.  AB  : :  sin.  B  :  sin.  ACB=122°  11'  6". 
Ex.  2.  In  the  spherical  triangle  ABC,  the  side  AB=40°  35', 
AC =39°  10',  and  BC=71°  15 .     Required  the  angles. 

[  A=130°  35'  66", 

Ans.  )  B=  30°  25'  34". 

(  C=  31°  26'  32". 

Case  VI. 

(224.)  Given  the  three  angles  of  a  spherical  triangle^  to 
find  the  sides. 

If  A,  B,  C  are  the  angles  of  the  given  triangle,  and  a,  b,  c 
its  sides,  then  180°-A,  180°-B,  and  180°— C  are  the  side** 


Spherical    Trigonometry. 


171 


of  its  polar  triangle,  whose  anglet  may  be  found  by  Case  Y. 
Then  the  supplements  of  those  angles  will  be  the  sides  a,  &,  c 
of  the  proposed  triangle. 

Ex.  1.  In  the  spherical  triangle  ABC,  the  angle  A=125° 
34',  B=98°  44',  and  C=61°  53'.     Required  the  sides. 
The  sides  of  the  polar  triangle  are 

54°  26',  81°  16',  and  118°  7'. 
From  which,  by  Case  V.,  the  angles  are  found  to  be 

134°  6'  21",  41°  28'  17",  and  53°  34'  47". 
Hence  the  sides  of  the  proposed  triangle  are 
AB=45°  53'  39",  BC=138°  31'  43",  and  AC=126°  25'  13" 

Ex.  2.  In  the  spherical  triangle  ABC,  the  angle  A=109° 
55',  B=116°  38',  and  C=120°  43'.     Required  the  sides. 

Ca=  98°  21' 20". 

il^5.  ]Z>=109°50'  10". 

(c -115°  13'    1" 


TRIGONOMETRICAL  FORMULAE. 

(225.)  Let  ABC  be  any  spherical 
triangle,  and  from  the  angle  B  draw 
the  arc  BD  perpendicular  to  the  base 
AC.  Represent  the  sides  of  the  trian- 
gle by  a,  Z>,  c,  and  the  segment  AD  by 
X ;  then  will  CD  be  equal  to  b—x. 
By  Theorem  III.,  Cor.,  1, 

cos.  c  :  cos.  a  :  :  cos.  x  :  cos.  (b—x) 

COS.  b  COS.  re + sin.  b  sin.  x, 


cos.  X 


R 


(Trig.,  Art.  72),  formula  (4). 

Whence 

R  cos.  a  cos.  a;=cos.  b  cos.  c  cos.  rc+sin.  b  cos.  c  sin.  x; 
or,  dividing  each  term  by  cos.  x,  and  substituting  the  value  o« 

(Art.  28),  we  obtain 


cos.  X 


R"*  COS.  a=R  COS.  b  cos.  c+sin.  b  cos.  c  tan.  x. 


But  by  Theorem  II.,  Cor.  2,  we  have 


tan.  x= 


R  COS.  A    COS.  A  sin.  c 


cot.  c 


COS.  c 


(Art.  28). 


172  •         Trigonometry. 

Hence   R'  cos.  a=R  cos.  b  cos.  c+sin.  b  sin.  c  cos.  A,    (1^ 
from  which  all  the  formulse  necessary  for  the  solution  of  spheri- 
cal triangles  may  be  deduced. 
In  a  similar  manner  we  obtain 

R"  COS.  6=R  cos.  a  cos.  c+sin.  a  sin.  c  cos.  B,         (2) 
R'  COS.  c=R  COS.  a  cos.  6+sin.  a  sin.  b  cos.  C.         (3) 
These  equations  express  the  following  Theorem : 
The  square  of  radius  multiplied  by  the  cosine  of  either  side 
of  a  spherical  triangle,  is  equal  to  radius  into  the  product  of 
the  cosines  of  the  two  other  sides,  plus  the  product  of  the  sines 
of  those  sides  into  the  cosine  of  their  included  angle. 
(226.)  From  equation  (1)  we  obtain,  by  transposition, 
R'*  COS.  a— R  cos.  b  cos.  c 

COS.  A= : ; : , 

sm.  b  sm.  c 
a  formula  which  furnishes  an  angle  of  a  triangle  when  the 
three  sides  are  known. 

If  we  add  R  to  each  member  of  this  equation,  we  shall  have 
R'  COS.  a+R  sin.  b  sin.  c— R  cos.  b  cos.  c 


R+cos.  A: 


sin.  b  sin.  c 
2  cos.  '^k 


But,  by  Art.  74,  R+cos.  A=-       ^ 

And,  by  Art.  72,  formula  (2),  by  transposition, 

R  sin.  b  sin.  c— R  cos.  b  cos.  c=  — R'  cos.  (6+c). 

Hence,  by  substitution,  we  obtain 

2  cos.  ''JA_R''(cos.  a— COS.  {b+c)) 

R        ~  sin.  b  sin.  c 

_2R  sin.  ^(a+b+c)  sin.  \{b-\-c—a) 

sin.  b  sin.  c  ' 

by  Art.  75,  formula  (4). 

If,  then,  we  put  s=^(a+b+c),  that  is,  half  the  sum  of  the 

sides,  we  shall  find 

1  A     T.4  /sin.  s  sin.  (s  —  a) 
COS.  JA=R\/ -. — r—r^ -.  (4) 

^       sm.  b  sm.  c  ^  ' 

By  subtracting  cos.  A  from  R  instead  of  adding,  we  shall 
obtain,  in  a  similar  manner, 

•      lA     T».  /sin.  (s-b)  sin.  {s-c) 

sm.  iA=RV ^ — J—. — ^ -,  (5) 

^  ^  sm.  b  sm.  c  ^  ' 


Spherical    Trigonometry.  173 

Either  formula  (4)  or  (5)'  may  be  employed  to  compute  the 
angles  of  a  spherical  triangle  when  the  three  sides  are  known, 
and  this  method  may  be  preferred  to  that  of  Art.  223. 

Ex.  1.  In  a  spherical  triangle  there  are  given  a—6S°  50', 
^>=80°  19',  and  c=120°  47'.     Required  the  three  angles. 
Here  half  the  sum  of  the  sides  is  132°  28' =5. 
Also,  s-a=68°  38'! 

Using  formula  (4),  we  have 

log.  sine  5,  132°  28'     .     .     9.867862 

log.  sine  (s-a),  68°  38'  .  .  9.969075 
-log.  sine  b,  80°  19'  comp.  0.006232 
-log.  sine  c,       120°  47'  comp.    0.065952 

Sum     19.909121 
log.  COS.  JA,         25°  45'  19"        9.954560. 
Hence  the  angle  A=51°  30'  38". 

The  remaining  angles  may  be  found  by  Theorem  HI.,  or  by 
formulas  similar  to  formula  (4). 

i-D    -DA  /sin.  s  sin.  (s—b) 

COS.  AB  =  R\/ : r-^— — -, 

^       sm.  a  sm.  c 

,^    _,    /sin.  s  sin.  (s—c) 

COS.  A-C=RV : -      2.     ' 

^       sm.  a  sm.  o 

We  thus  find  the  angle  B=  59°  16'  46", 

and  C=131°  28'  36". 

Ex.  2.  In  a  spherical  triangle  there  are  given  ^^=115°  20', 

6=57°  30',  and  c=82°  28'.     Required  the  three  angles. 

(  A=126°  35'    2". 

Ans.  ]  B=  48°  31'  42". 

(  C=:  61°  43'  58". 

(227.)  By  means  of  the  polar  triangle,  we  may  convert  the 

preceding  formulae  for  angles  into  formulae  for  the  sides  of  a 

triangle,  since  the  angles  of  every  triangle  are  the  supplements 

of  the  sides  of  its  polar  triangle.     Let,  then,  a',  b',  c',  A',  B', 

C  represent  the  sides  and  angles  of  the  polar  triangle,  and  wo 

shall  have 

A=180°-a',  B=180°-6',  C=180°-c; 

a=180°-A',  6=180°-B',  c=180°-C' 

Therefore        sin.  iA=sin.  (90°— ^a')  =cos.  ^a', 

cos.  ^A=cos.  (90°— ^a')  =sin.  ^a', 


174  Trigonometry. 

sin.  ^=sin.  vl80°-B')=sin.  B', 
sin.  c=sin.  (180°-C')=sin.  C. 
Also,  if  we  put  S'=half  the  sum  of  the  angles  of  the  polaf 
triangle,  we  shall  have 

a+6+c=540°-(A'+B'+C'), 
or  5=270° -S', 

whence  sin.  5=— cos.  S', 

sin.  (5-a)=sin.  [90°-(S'-A')]=cos.  (S'-A'^ 
sin.  (s-b)=cos.  (S'-B'), 
sin.  (5-c)=cos.  (S'-C). 
By  substituting  these  values  in  formula  (5),  Art.  226,  and 
omitting  all  the  accents,  since  the  equations  are  applicable  to 
any  triangle,  we  obtain 


,       ^     /cos.  (S-B)cos.(S-C) 

^  sm.  B  sm.  C  '  ^  ' 

and  formula  (4)  becomes 

•      1       -n.  /-COS.  Scos.  (S-A)  _ 

sm.  J«j=R V : — -D    .  ^  ^ -,  (7) 

^  ^         sm.  B  sm.  C        '  ^  ' 

which  formuloB  enable  us  to  compute  the  sides  of  a  triangle 
when  the  three  angles  are  known ;  and  this  method  may  be 
preferred  to  that  of  Art.  224. 

In  a  similar  manner,  by  means  of  the  polar  triangle,  we 
derive  from  formula  (1),  Art.  225,  the  equation 

R'  COS.  A=cos.  a  sin.  B  sin.  C— R  cos.  B  cos.  C ;     (8) 

that  is,  the  square  of  radius  multiplied  by  the  cosine  of  either 
angle  of  a  spherical  triangle,  is  equal  to  the  product  of  the 
sines  of  the  two  other  angles  into  the  cosine  of  their  included 
side,  minus  radius  into  the  product  of  their  cosines. 

Ex.  1.  In  a  spherical  triangle  ABC,  there  are  given  A= 
130°  30',  B=30°  50',  and  0=32°  5'.  Required  the  three 
sides. 

Here  half  the  sum  of  the  angles  is  96°  42  30"= S. 

Also,  S-A=-33°47'30", 

S-^B=     65°  52' 30", 
S_C=    64°  37' 30" 

Using  formula  (6),  we  have 


Spherical   Trigonometry.  175 

Jog.  COS.  (S-B),  e^""  52'  30"     .  9.611435 

log.  COS.  (S-C),  64°  37'  30"     .  9.631992 

-log.  sin.  B,       30°  50'  comp.  0.290270 

-log.  sin.  C,      32°    5'  comp.  0.274781 

Sum  19.808478 


log.  cos.  ^a,        36°  40'  1"  9.904239. 

Hence  the  side  a=73°  20'  2". 

The  remaining  sides  may  be  found  by  Theorem  III.,  or  by 
formulas  similar  to  formula  (6). 

,,     ^.  /cos.  (S-A)cos.  (S-C) 
'''•  ^'=^y  sin.  A  sin.  C  - 

COS.  ic^RV----(^-^)-.-B. 
^  sin.  A  sm.  B 

We  thus  find  the  side  6=40°  13°  12", 

and  c=42°    0'  12". 

Ex.  2.  In  the  spherical  triangle  ABC,  the  angle  A=129° 

30',  B=54°  35',  and  C=63°  5'.     Required  the  three  sides. 

C  «=120°  57'    5". 

Ans.  \b=  64°  55' 37" 

U=  82°  19'    0". 

(228.)  Formula  (1),  Art.  225,  will  also  furnish  a  new  test 

for  removing  the  ambiguity  of  the  solution  in  Case  I.  of  oblique 

angled  triangles.     For  we  have 

.     R'  cos.  «— R  cos.  b  COS.  c 

COS.  A  = ; — J — ; . 

sm.  0  sin.  c 

Now  if  cos.  a  is  greater  than  cos.  6,  we  shall  have 
R'  COS.  a>R  COS.  b  cos.  c, 
or  the  sign  of  the  second  member  of  the  equation  will  be  the 
same  as  that  of  cos.  a,  since  the  denominator  is  necessarily 
positive,  and  cos.  c  is  less  than  radius.  Hence  cos.  A  and  cos. 
a  will  have  the  same  sign ;  or  A  and  a  will  be  of  the  same 
species  when  cos  a>cos.  6,  or  sin.  a<sin.  b  ;  that  is. 

If  the  sine  of  the  side  opposite  to  the  required  angle  is  less 
than  the  sine  of  the  other  given  side,  there  will  be  but  one 
triangle. 

But  if  COS.  a  is  less  than  cos.  b,  then  such  a  value  may  be 
given  to  c  as  to  render 


176  Trigonometry. 

R'^  COS.  a<R  COS.  h  cos.  c, 
or  the  sign  of  the  second  member  of  the  equation  will  depend 
upon  the  value  of  cos.  c  ;  that  is,  c  may  be  taken  so  as  to  ren- 
der cos.  A  either  positive  or  negative.     Hence 

If  the  sine  of  the  side  opposite  to  the  required  angle  is 
greater  than  the  sine  of  the  other  given  side,  there  will  he 
two  triangles  which  fulfill  the  given  conditions. 

(229.)  Formula  (8),  Art.  227,  wijl  furnish  a  trst  for  re- 
moving  the  ambiguity  in  Case  II.  of  oblique-angled  triangles. 
For  we  have 

R'*  COS.  A+R  COS.  B  COS.  C 

cos.  a-= : — :5— ; — ; 

sm.  B  sm.  C 

trom  which  it  follows,  as  in  the  preceding  article,  that  if  cos.  A 
is  greater  than  cos.  B,  A  and  a  will  be  of  the  same  species.  But 
if  COS.  A  is  less  than  cos.  B,  then  such  values  may  be  given 
to  C  as  to  render  cos.  a  either  positive  or  negative.     Hence 

If  the  sine  of  the  angle  opposite  to  the  required  side  is  less 
than  the  sine  of  the  other  given  angle,  there  will  be  but  one 
triangle  ; 

But,  if  the  sine  of  the  angle  opposite  to  the  required  side 
is  greater  than  the  sine  of  the  other  given  angle,  there  will 
be  two  triangles  which  fulfill  the  given  conditions. 

SAILING  ON  AN  ARC  OF  A  GREAT  CIRCLE. 

(230.)  It  is  demonstrated  in  Geom.,  Prop.  3,  B.  IX.,  that  the 
shortest  path  from  one  point  to  another  on  the  surface  of  a 
sphere  is  the  arc  of  a  great  circle  which  joins  the  two  given 
points.  Hence,  if  it  is  desired  to  sail  from  one  port  to  another 
by  the  shortest  route,  it  is  necessary  to  follow  an  arc  of  a  great 
circle,  and  this  arc  generally  does  not  coincide  with  a  rhumb- 
line. 

The  bearing  and  distance  from  one  place  to  another  on  the 
arc  of  a  great  circle  may  be  computed  from  the  latitudes  and 
longitudes  of  the  places  by  means  of  Spherical  Trigonometry 

Thus,  let  P  be  the  pole  of  the  earth,  EQ,  a  part  of  the  equa- 
tor, and  A  and  B  the  two  given  places  comprehended  between 
the  meridians  PE  and  PQ,.  Then  PA  is  the  complement  of 
the  latitude  of  A,  PB  is  the  complement  of  the  latitude  of  B, 


Spherical   Trigonometry. 


1/7 


and  the  angle  P  is  measured  by  the  arc  EQ,,  which  is  the 
difference  of  longitude  bfitween  the  two 
places.  Hence,  in  the  triangle  ABP,  we 
have  given  two  sides  AP,  BP,  and  the  in- 
cluded angle  P,  from  which  we  may  com- 
pute the  side  AB,  and  the  angles  A  and  B, 
according  to  Case  III.  of  oblique-angled  tri- 
angles. 

Ex.  1.  Required  the  course  and  distance 
from  Nantucket  Shoals,  in  latitude  41°  4' 
N.,  longitude  69°  55'  "VV.,  to  Cape  Clear,  in 
latitude  51°  26 
circle. 

Here  we  have  given 

the  angle  P=69°  55'-  9°  29'=60°  26' ; 
the  side  PA=90°  -41°  4'=48°  56' ; 
the  side  PB=90°       -51°  26'=38°  84'. 

Then  cot.  PB  :  cos.  P  : :  R  :  tan.  PD=21°  28'  35". 

Whence  AI)=27°  27'  25". 

Also  sin.  AD  :  sin.  PD  :  :  tan.  P  :  tan.  A   =54°  27' 
and        sin.  A     :  sin.  PB  :  :  sin.  P  :  sin.  AB=41°  47' 

41°  47'  28"  is  equal  to  2507.47  nautical  miles. 

Hence  the  course  from  Nantucket 
Shoals  to  Cape  Clear  is  N.  54°  27' 
E.,  and  the  distance  is  2507.47  miles. 


N.,  longitude  9°  29'  W.,  on  the  arc  of  a  gr«flt 


21 

28' 


According 


to   Me'rcator' 


sailing. 


the  course  on  a  rhumb-line,  found  on 
page  152,  is  N.  76°  E.,  and  the  dis- 
tance 2572.9  miles.  Hence  the  dis- 
tance on  an  arc  of  a  great  circle  is  65.4  miles  less  than  on  a 
rhumb-line,  and  the  former  course  is  21|^  degrees  more  north- 
erly than  the  latter. 

While  sailing  on  a  rhumb-line  the  course  of  a  ship  remains 
always  the  same,  but  while  sailing  on  an  arc  of  a  great  circle 
the  course  is  continually  changing.  The  preceding  course  is 
that  with  which  the  ship  starts  from  Nantucket,  and  a  new 
computation  of  the  course  should  be  made  every  day  or  two ;  or 
.t  might  be  more  convenient  to  compute  beforehand  the  position 
of  the  points  in  which  the  great  circle  intersects  the  meridians 


178  Trigonometry. 

for  every  five  degrees  of  longitude,  and  the  ship  niiglit  bt; 
steered  upon  a  direct  course  for  these  points  successively. 

Ex.  2.  Required  the  course  ana  distance  from  Nantucket 
Shoals  to  G-ibraltar,  in  latitude  36°  6'  N.,  longitude  5°  2i^  "^  , 
on  the  shortest  route. 

Ans.  The  course  is  N.  73°  29'  E 
Distance  2974.1  miles. 
Ex.  3.  Required  the  course  and  distance  from  Sandy  Hook, 
in  latitude  40°  28'  N.,  longitude  74°  1'  "W.,  to  Madeira,  in 
latitude  32°  38'  N.,  longitude  16°  55'  W.,  on  the  shortest  route 

Ans.  The  course  is  N.  80°  53'  E. 
Distance  2744.1  miles. 
Ex.  4.  Required  the  course  and  distance  from  Sandy  Hook 
to  St.  Jago,  in  latitude  14°  54'  N.,  longitude  23°  30'  W.,  on 
tJie  shortest  route. 

Ans.  The  course  is  S.  74°  46'  E. 
Distance  3037.6  miles. 
Ex.  5.  Required  the  course  and  distance  from  Sandy  Hook 
to  the  Cape  of  Good  Hope,  in  latitude  34°  22'  S.,  longitude 
18°  30'  E.,  on  the  shortest  route. 

A71S.  The  course  is  S.  63°  48'  B 
Distance  6792  miles. 


EXAMPLES  FOR  PEACTICE. 

PLANE  TRIGONOMETRY. 

Prob.  1.  G-iven  the  three  sides  of  a  triangle,  627,  718.9,  and 
1140,  to  find  the  angles. 

Ans.  29°  44^  2^",  34°  39^  26^^  and  115°  36^  32^^ 
Prob.  2.  In  the  triangle  ABC,  the  angle  A  is  given  89°  45^ 
43^^,  the  side  AB  654,  and  the  side  AC  460,  to  find  the  remain- 
ing parts. 

Ans.  BC  =  798;  the  angle  B  =  35°  12^  1^  and  the  angle 

C=:55o2'16^^ 
Prob.  3.  In  the  triangle  ABC,  the  angle  A  is  given  6Q°  12' 
45'^  the  side  BC  2597.84,  and  the  side  AC  3084.33,  to  find  the 
remaining  parts. 

Ans.  B  =  80°  39^  40"^  C  =  43°    1'  35^  c  =  2136.8 ; 
or,  Bz=99    20  20,    C=24    26  6d,    c=.  1293.8. 
Prob.  4.  In  the  triangle  ABC,  the  angle  A  is  given  44°  13' 
24'',  the  angle  B  55°  59'  58",  and  the  side  AC  368,  to  find  the 
remaining  parts. 

Ans.  C=:79°  46'  38",  AB  =  436.844,  and  BC==  309.595. 
Prob.  5,  In  a  right-angled  triangle,  if  the  sum  of  the  J^- 
pothenuse  and  base  be  3409  feet,  and  the  angle  at  the  base  53° 
12'  14",  what  is  the  perpendicular  ? 

Ans.  1707.2  feet. 
Prob.  6.  In  a  right-angled  triangle,  if  the  difference  of  the 
hypothenuse  and  base  be  169.9  yards,  and  the  angle  at  the  base 
42°  36'  12",  what  is  the  length  of  the  perpendicular  ? 

Ans.  435.732  yards. 
Prob.  7.  In  a  right-angled  triangle,  if  the  sum  of  the  base 
and  perpendicular  be  123.7  feet,  and  the  angle  at  the  base  58° 
19'  32",  what  is  the  length  of  the  hypothenuse  ? 

Ans,  89.889  feet. 
Prob.  8.  In  a  right-angled  triangle,  if  the  difference  of  the 
base  and  perpendicular  be  12  yards,  and  the  angle  at  the  base 
38°  1'  8",  what  is  the  length  of  the  hypothenuse  ? 

Ans.  69.81  yards. 


180  Trigonometry. 

Prob.  9.  A  May-pole,  oO  feet  11  inches  high,  at  a  certain 
time  will  cast  a  shadow  98  feet  6  inches  long ;  what,  then,  is 
the  breadth  of  a  river  which  runs  within  20  feet  6  inches  of  the 
foot  of  a  steeple  300  feet  8  inches  high,  if  the  steeple  at  the 
same  time  throws  its  shadow  30  feet  9  inches  beyond  the  stream  ? 

Ans.  530  feet  5  inches. 

Prob.  10.  A  ladder  40  feet  long  may  be  so  placed  that  it  shall 
reach  a  window  33  feet  from  the  ground  on  one  side  of  the 
street,  and  by  turning  it  over,  without  moving  the  foot  out  of 
its  place,  it  will  do  the  same  by  a  window  21  feet  high  on  the 
other  side.     Required  the  breadth  of  the  street. 

Ans.  56.649  feet. 

Prob.  11.  A  May-pole,  whose  top  was  broken  off  by  a  blast 
of  \vind,  struck  the  ground  at  the  distance  of  15  feet  from  the 
foot  of  the  pole ;  what  was  the  height  of  the  whole  May-pole, 
supposing  the  length  of  the  broken  piece  to  be  39  feet? 

Ans.  75  feet. 

Prob.  12.  How  must  three  trees,  A,  B,  C,  be  planted,  so  that 
the  angle  at  A  may  be  double  the  angle  at  B,  the  angle  at  B 
double  the  angle  at  C,  and  a  line  of  400  yards  may  just  go 
round  them  ? 

Ans.  AB= 79.225,  AC  =  142.758,  and  BC  =  178.017  yards. 

Prob.  13.  The  town  B  is  half  way  between  the  towns  A  and 
C,iftnd  the  towns  B,  C,  and  D  are  equidistant  from  each  other. 
What  is  the  ratio  of  the  distance  AB  to  AD  ? 

Ans.  As  unity  to  -^3. 

Prob.  14.  There  are  two  columns  left  standing  upright  in  the 
ruins  of  Persepolis ;  the  one  is  66  feet  above  the  plain,  and  the 
other  48.  In  a  straight  line  between  them  stands  an  ancient 
statue,  the  head  of  which  is  100  feet  from  the  summit  of  the 
higher,  and  84  feet  from  the  top  of  the  lower  column,  the  base 
of  which  measures  just  74  feet  to  the  centre  of  the  figure's  base. 
Required  the  distance  between  the  tops  of  the  two  columns. 

Ans.  156.68  feet. 

Prob.  15.  Prove  that  tang.  (45°-^^)=:^— ^4- 
^  "^  '     1  + tang.  6 

Prob.  16.  One  angle  of  a  triangle  is  45°,  and  the  perpendic- 
ular from  this  angle  upon  the  opposite  base  divides  the  base  into 
two  parts,  which  are  in  the  ratio  of  2  to  3.     What  are  the 


Examples   for   Practice.  181 

parts  into  which  the  vertical  angle  is  divided  by  this  perpen- 
dicular ? 

Ans.  18°  26^  6''  and  26°  33^  54:''. 
Prob.  17.  Prove  that  sin.  36  =  3  sin.  &— 4  sin.^  b. 
Prob.  18.  One  side  of  a  triangle  is  25,  another  is  22,  and  the 
angle  contained  by  these  two  sides  is  half  that  of  the  angle  op- 
posite the  side  25.     What  is  the  Value  of  the  included  angle? 

Ans.  39°  58^  51^^ 
Prob.  19.  One  side  of  a  triangle  is  25,  another  is  22,  and 
the  angle  contained  by  these  two  sides  is  half  that  of  the  angle 
opposite  the  side  22.     What  is  the  value  of  the  included  angle  ? 

Ans.  30°  W  38^^ 
Prob.  20.  Two  sides  of  a  triangle  are  in  the  ratio  of  11  to  9, 
and  the  opposite  angles  have  the  ratio  of  3  to  1.     What  are 
those  angles  ? 

Ans.  The  sine  of  the  smaller  of  the  two  angles  is  -|,  and  of 
the  greater  |f ;  the  angles  are  41°  48'  37'"  and 
125°  25'  51''. 
Prob.  21.  One  side  of  a  triangle  is  15,  and  the  difference  of 
the  two  other  sides  is  6 ;  also,  the  angle  included  between  the 
first  side  and  the  greater  of  the  two  others  is  60°.     What  is  the 
length  of  the  side  opposite  to  this  angle  ? 

Ans.  57. 
Prob.  22.  One  side  of  a  triangle  is  15,  and  the  difference  of 
the  two  other  sides  is  6 ;  also,  the  angle  opposite  to  the  greater 
of  the  two  latter  sides  is  60°.     What  is  the  length  of  said  side  ? 

Ans.  13. 
Prob.  23.  One  side  of  a  triangle  is  15,  and  the  opposite  an- 
gle is  60°  ;  ^Iso,  the  difference  of  the  two  other   sides  is  6. 
What  are  the  lengths  of  those  sides  ? 

Ans.  11.0712  and  17.0712. 
Prob.  24.  The  perimeter  of  a  triangle  is  100  ;  the  perpendic- 
ular let  fall  from  one  of  the  angles  upon  the  opposite  base  is  30, 
and  the  angle  at  one  end  of  this  base  is  50°.     What  is  the 
length  of  the  base  ? 

Ans.  30.388. 


182  Trigonometry.  . 

MENSUKATION  OF  SURFACES  AND  SOLmS. 

Prob.  1.  The  base  of  a  triangle  is  20  feet,  and  its  altitude 
18  feet.  It  is  required  to  draw  a  line  parallel  to  the  base  so  as 
to  cut  off  a  trapezoid  containing  80  square  feet.  What  is  the 
length  of  the  line  of  section,  and  its  distance  from  the  base  of 
the  triangle? 

Ans.  Length  14.907  feet ;  distance  from  base  4.584  feet. 

Prob.  2.  The  base  of  a  triangle  is  20  feet,  one  angle  at  the 
base  is  63°  26^,  and  the  other  angle  at  the  base  is  56°  19^  It 
is  required  to  draw  a  line  parallel  to  the  base,  so  as  to  cut  off  a 
trapezoid  containing  109  square  feet.  What  is  the  length  of 
the  line  of  section,  and  its  distance  from  the  base  of  the  tri- 
angle ? 

Ans.  Length  12.070  feet ;  distance  from  base  6.797  feet. 

Prob.  3.  In  a  perpendicular  section  of  a  ditch,  the  breadth  at 
the  top  is  26  feet,  the  slopes  of  the  sides  are  each  45°,  and  the 
area  140  square  feet.  Required  the  breadth  at  bottom  and  the 
depth  of  the  ditch. 

Ans.  Breadth  10.77  feet ;  depth  7.615  feet. 

Prob.  4.  The  altitude  of  a  trapezoid  is  23  feet ;  the  two  par- 
allel sides  are  76  and  36  feet ;  it  is  required  to  draw  a  line  par- 
allel to  the  parallel  sides,  so  as  to  cut  off  from  the  smaller  end 
of  the  trapezoid  a  part  containing  560  square  feet.  What 
is  the  length  of  the  line  of  section,  and  its  distance  from  the 
shorter  of  the  two  parallel  sides  ? 

Ans.  Length  56.954  feet;  distance  12.048  feet. 

Prob.  5.  From  the  greater  end  of  a  trapezoidal  field  whose 
parallel  ends  and  breadth  measure  12,  8,  and  10^  chains  re- 
spectively, it  is  required  to  cut  off  an  area  of  six  acres  by  a 
fence  parallel  to  the  parallel  sides  of  the  field.  What  is  the 
length  of  the  fence,  and  its  distance  from  the  greater  side. 

Ans.  Length  of  fence  9.914  chains ;  distance  from  greater 
side  5.476  chains. 

Prob.  6.  There  are  three  circles  whose  radii  are  20,  28,  and 
29  inches  respectively.  Required  the  radius  of  a  fourth  circle, 
whose  area  is  equal  to  the  sum  of  the  areas  of  the  other  three. 

Ans.  45  inches. 

Prob.  7.  In  constructing  a  rail-road,  the  pathway  of  which 


Examples   for   Practice.  183 

•is  24  feet  broad,  it  is  necessary  to  make  a  cutting  40  feet  in 
depth;  what  must  be  the  breadth  of  the  cutting  at  top,  sup- 
posing the  slopes  of  the.  sides  to  be  65°  ? 

Ans.  61.805  feet. 
Prob.  8.  The  sides  of  a  quadrilateral  field  are  690  yards, 
467  yards,  359  yards,  and  428  yards ;  also,  the  angle  contained 
between  the  first  and  second  sides  is  57°  30',  and  the  angle  be- 
tween the  third  and  fourth  sides  122°  30'.  Required  the  area 
of  the  field. 

Ans.  200677.2  square  yards. 
Prob.  9.  There  are  two  regular  pentagons,  one  inscribed  in  a 
circle,  and  the  other  described  about  it ;  and  the  difference  of 
the  areas  of  the  pentagons  is  100  square  inches.     Required  the 
radius  of  the  circle. 

Ans.  8.926  inches. 
Prob.  10.  What  is  the  length  of  a  chord  cutting  off  one  third 
part  of  a  circle,  whose  diameter  is  289  feet. 

Ans.  278.67  feet. 
Prob.  11.  The  area  of  a  triangle  is  1012 ;  the  length  of  the 
side  a  is  to  that  of  6  as  4  to  3,  and  c  is  to  ^  as  3  to  2.     Re- 
quired the  length  of  the  sides. 

Ans.  a  =  52.470,  ^  =  39.353,  c= 59.029. 
Prob.  12.  The  area  of  a  trianglo  is  144,  the  base  is  24,  and 
one  of  the  angles  at  the  base  is  30^.     Required  the  other  sides 
of  the  triangle. 

Ans.  24  p.nd  12.4233. 
Prob.  13.  Seven  men  bought  a  grinding-stone  of  60  inches 
diameter,  each  paying  one  seventh  part  of  the  expense.     What 
part  of  the  diameter  must  each  grind  down  for  his  share  ? 
Ans.  The  1st,  4.4508  inches ;  2d,  4.8400  inches ;  3d,  5.3535 
inches ;  4th,  6^765  inches ;  5th,  7.2079  inches ; 
6th,  9.3935  inches;  7th,  22.6778  inches. 
Prob.  14.  The  area  of  an  equilateral  triangle  is  17  square 
feet  and  83  square  inches.     What  is  the  length  of  each  side  ? 

Ans.  76.45  inches. 
P7'ob,  15*.  The  parallel  sides  of  a  trapezoid  are  20  and  12  feet, 
and  the  other  sides  are  15  and  17  feet.     Required  the  area  of 
the  trapezoid. 

Ans.  240  square  feet. 


1^  Trigonometry. 

Prob.  16.  How  many  square  yards  of  canvas  are  required  to 
make  a  conical  tent  which  is  20  feet  in  diameter  and  12  feet 
high? 

Ans.  54.526  square  yards. 
Prob.  17.  The  circumference  of  an  hexagonal  pillar  is  7  feet, 
and  the  height  11  feet  2  inches.    Required  the  solid  contents  of 
the  pillar. 

Ans.  39.488  cubic  feet. 
Prob.  18.  The  base  of  the  great  pyramid  of  Egypt  is  a  square 
whose  side  measures  746  feet,  and  the  altitude  of  the  pyramid 
is  450  feet.     Required  the  volume  of  the  pyramid. 

Ans.  83,477,400  cubic  feet. 
Prob.  19.  A  side  of  the  base  of  a  frustum  of  a  square  pyra- 
mid is  25  inches,  a  side  of  the  top  is  9  inches,  and  the  height  is 
20  feet.     Required  the  volume  of  the  frustum. 

Ans.  43.102  cubic  feet. 
Prob.  20.  Three  persons,  having  bought  a  sugar-loaf,  would 
divide  it  equally  among  them  by  sections  parallel  to  the  base. 
It  is  required  to  find  the  altitude  of  each  person's  share,  suppos- 
ing the  loaf  to  be  a  cone  whose  height  is  20  inches. 

Ans.  13.8672,  3.6044,  and  2.5284  inches. 
Prob.  21.  If  a  cubical  foot  of  brass  were  to  be  drawn  into 
wire  of  one  thirtieth  of  an  inch  in  diameter,  it  is  required  to  de- 
termine the  length  of  the  said  wire,  allowing  no  loss  in  the  metal. 
Ans.  55003.94  yards ;  or  31  miles  443.94  yards. 
Prob.  22.  How  high  above  the  surface  of  the  earth  must  a 
person  be  raised  to  see  one  third  of  its  surface  ? 

Ans.  The  height  of  its  diameter. 
Prob.  23.  If  a  heavy  sphere,  whose  diameter  is  4  inches,  b'e 
let  fall  into  a  conical  glass  full  of  water,  whose  diameter  is  5, 
and  altitude  6  inches,  it  is  required  to|Jetermine  how  much  wa- 
ter will  run  over. 

Ans.  26.272  cubic  inches. 
Prob.  24.  The  capacity  of  a  cylinder  is  a  cubic  feet,  and  its 
convex  surface  is  b  square  feet.     Required  the  dimensions  of 
the  cylinder. 

Ans.  Radius  of  base=— -,  and  altitude = — . 
b  4arr 

Prob.  25.  'A  triangular  pyramid,  the  sides  of  whose  base  are 


Examples   for   Practice.  185 

13,  14,  and  15  inches  respectively,  and  whose  altitude  is  16 
inches,  is  cut,  at  the  distance  of  2  inches  from  the  vertex,  by  a 
plane  parallel  to  the  base.  Required  the  volume  of  the  frustum 
of  the  pyramid. 

Ans.  447.125  cubic  inches. 
Prob.  26.  The  altitude  of  a  cone  is  10  inches,  and  the  radius 
of  its  base  is  5  inches.    At  what  distance  from  the  base  must  a 
plane  pass  parallel  to  the  base,  so  as  to  cut  off  a  frustum  whose 
capacity  is  20  cubic  inches  ? 

Ans.  0.2614  inches. 

# 

SURVEYING. 

Prob.  1.  The  angle  of  elevation  of  a  spire  I  found  to  be  39° 
27^,  and  going  directly  from  it  225  feet  on  a  horizontal  plane,  I 
found  the  angle  to  be  only  24°  38^.     What  is  the  height  of  the 
spire,  and  the  distance  from  its  base  to  the  second  station  ? 
Ans.  Height  238.02  feet,  distance  508.18  feet. 

Prob.  2.  Wishing  to  know  the  distance  of  an  inaccessible  ob- 
ject, I  measured  a  horizontal  base-line  1328  feet,  and  found  the 
angles  at  the  ends  of  this  line  were  84°  23''  and  43°  19^    What 
was  the  distance  of  the  object  from  each  end  of  the  base-line  ? 
Ans.  1151.44  feet,  and  1670.35  feet. 

Prob.  3.  Wishing  to  know  the  distance  between  two  inacces- 
sible objects,  C  and  D,  I  measured  a  base-line,  AB,  3784  feet, 
and  found  the  angle  BAD =47°  32^  the  angle  DAC  =  3£)°  53^, 
the  angle  ABC  =46°  34^  and  the  angle  CBD  =  38°  V.  What 
is  the  distance  from  C  to  D  ? 

Ans.  3257.36  feet. 

Prob.  4.  Suppose  a  light-house  built  on  the  top  of  a  rock ;  the 
distance  between  the  place  of  observation  and  that  part  of  the 
rock  which  is  level  with  the  eye,  and  directly  under  the  build- 
ing, is  1860  feet ;  the  distance  from  the  top  of  the  rock  to  the 
place  of  observation  is  2538  feet,  and  from  the  top  of  the  build- 
ing 2550  feet.     Required  the  height  of  the  light-house. 

Ans.  17  feet  7  inches. 

Prob.  5.  At  85  feet  distance  from  the  bottom  of  a  tower, 
standing  on  a  horizontal  plane,  the  angle  of  its  elevation  was 
found  to  be  52°  30^     Required  the  altitude  of  the  tower. 

Ans.  llOi  feet. 


186  Trigonometry. 

Prob,  6.  At  a  certain  station,  the  angle  of  elevation  of  an  in- 
accessible tower  was  26°  30^ ;  but,  measuring  225  feet  in  a  di- 
rect line  toward  it,  the  angle  was  then  found  to  be  51°  30^ 
Required  the  height  of  the  tower,  and  its  distance  from  the  last 
station.  Ans.  Height  186  feet,  distance  147  feet. 

Proh.  7.  To  find  the  distance  of  an  inaccessible  castle  gate,  I 
measured  a  line  of  73  yards,  and  at  each  end  of  it  took  the  an- 
gle of  position  of  the  object  and  the  other  end,  and  found  the  one 
to  be  90°,  and  \h<d  other  61°  45^  Required  the  distance  of  the 
castle  from  each  station. 

Ans.  135.8  yards,  and  154.2  yai^s. 

Proh,  8.  From  the  top  of  a  tow^er  143  feet  high,  by  the  sea- 
side, I  observed  that  the  angle  of  depression  of  a  boat  was  35°. 
What  was  its  distance  from  the  bottom  of  the  tower  ? 

Ans.  204.22  feet. 

Prob.  9.  I  wanted  to  know  the  distance,  between  two  places, 
A  and  B,  but  could  not  meet  with  any  station  from  whence  I 
could  see  both  objects.  I  measured  a  line  CD  =  200  yards ;  from 
C  the  object  A  was  visibly,  and  from  D  the  object  B  was  visi- 
ble, at  each  of  which  places  I  set  up  a  pole.  I  also  measured 
FC  =  200  yards,  and  DE=200  yards,  and  at  F  and  E  set  up 
poles.  I  then  measured  the  angle  AFC  =  83o,  ACF  =  54°  31^ 
ACD==53^  30^  BDC  =  156°  25^  BDE=:54°  30^  and  BED  = 
88°  30^     Required  the  distance  from  A  to  B. 

«  Ans.  345.5  yards. 

Proh,  10.  From  the  top  of  a  light-house,  the  angle  of  depres- 
sion of  a  ship  at  anchor  was  3°  38^,  and  at  the  bottom  of  the 
light-house  the  angle  of  depression  was  2°  43^  Required  the 
horizontal  distance  of  the  vessel,  and  the  height  of  the  promon- 
tory above  the  level  of  the  sea,  the  light-house  being  85  feet 
high.  Ans.  Distance  5296.4  feet,  height  251.3  feet. 

Proh,  11.  An  observer,  seeing  a  cloud  in  the  west,  measured 
its  angle  of  elevation,  and  found  it  to  be  64°.  A  second  observ- 
er, situated  half  a  mile  due  east  from  the  first  station,  and  on 
the  same  horizontal  plane,  found  its  angle  of  elevation  at  the 
same  moment  of  time  to  be  only  35°.  Required  the  perpendic- 
ular height  of  the  cloud,  and  its  distance  from  each  observer. 

Ans.  Perpendicular  height  935.75  yards,  distances  1041.1 
and  1631.4  yards. 


Examples   for   Practice.  187 

Prob.  12.  An  observer,  seeing  a  balloon  in  the  north,  meas- 
ured its  angle  of  elevation,  and  found  it  to  be  36°  52^  A  second 
observer,  situated  one  mile  due  south  from  the  first  station,  and 
on  the  same  horizontal  plane,  found  its  angle  of  elevation  at  the 
same  instant  to  be  only  30°  58^  Required  the  perpendicular 
height  of  the  balloon,  and  its  distance  from  each  observer. 
Arts.  Perpendicular  height  3.003  miles,  distances  5.006 

and  5.837  miles. 
Proh.  13.  From  a  window  near  the  bottom  of  a  house  which 
seemed  to  be  on  a  level  with  the  bottom  of  a  steeple,  I  found  the 
angle  of  elevation  of  the  top  of  the  steeple  to  be  40°  ;  then  from 
another  window,  21  feet  directly  above  the  former,  the  like  angle 
was  37°  30^  What  was  the  height  and  distance  of  the  stee- 
ple ?  Ans,  Height  245.51  feet,  distance  292.59  feet 

P^ob.  14.  Wanting  to  know  my  distance  from  an  inaccessi- 
ble object,  P,  on  the  other  side  of  a  river,  and  having  no  instru- 
ment for  taking  angles,  but  a  chain  for  measuring  distances, 
from  each  of  two  stations,  A  and  B,  which  were  taken  at  300 
yards  asunder,  I  measured  in  a  direct  line  from  the  object  P  60^ 
yards,  viz.,  AC  and  BD  each  equal  to  60  yards ;  also,  the  diag- 
onal AD  measured  330  yards,  and  the  diagonal  BC  336  yards. 
What  was  the  distance  of  the  object  P  from  each  station  A 
and  B  ?  Ans.  AP  =  321.76  yards,  BP:=  300.09  yards. 

Prob.  15.  Having  at  a  certain  (unknown)  distance  taken  the 
angle  of  elevation  of  a  steeple,  I  advanced  60  yards  nearer  on 
level  ground,  and  then  observed  the  angle  of  elevation  to  be  the 
complement  of  the  former.  Advancing  20  yards  still  nearer, 
the  angle  of  elevation  now  appeared  to  be  just  double  of  the 
first.     Required  the  altitude  of  the  steeple. 

Ans,  74.162  yards. 
Prob.  16.  In  a  garrison  there  are  three 
remarkable  objects.  A,  B,  C,  whose  dis- 
tances from  each  other  are  known  to  be, 
AB  213,  AC  424,  and  BC  262  yards.  I 
am  desirous  of  knowing  my  position  and 
distance  at  a  station,  P,  from  which  I  ob- 
served the  angle  APB,  13°  30',  and  the 

J,- angle  CPB,  29°  50^ 

Ans.  AP  =  605.7122,  BPz=  429.6814,  CP= 524.2365. 


188  Trigonometry. 

Proh.  17.  Supposing  the  object  B  to  "be  on  the  opposite  side 
of  the  line  AC  (see  figure  to  Prob.  16),  and  that  the  distances  of 
the  pbjects  were,  AB  =  8  miles,  AC  =  12  miles,  and  BC  =  7|^  miles ; 
also,  the  angle  APB=rl9°,  and  the  angle  CPB=25°.  It  is  re- 
quired to  find  the  distances  AP,  BP,  and  CP. 

Ans.  AP  =  9.4711  miles,  BP=r  16.3369  miles, 
CP  =  16.8485  miles. 

Proh.  18.  In  a  pentangular  field,  beginning  with  the  south 
side,  and  measuring  round  toward  the  east,  the  first  or  south 
side  was  27.35  chains,  the  second  31.15  chains,  the  third  23.70 
chains,  the  fourth  29.25  chains,  and  the  fifth  22.20  chains; 
also,  the  diagonal  from  the  first  angle  to  the  third  was  38.00 
chains,  and  that  from  the  third  to  the  fifth  was  40.10  chains. 
Required  the  area  of  the  field. 

Ans.  117  A.  2  R.  39  P. 

Prob.  19.  The  following  are  the  dimensions  of  a  five-sided 
field,  ABCDE:  the  side  AB=  19.40  chains,  and  the  angle  B 
110°  30^;  the  side  BC=:  15.55  chains,  and  the  angle  C  117° 
45^ ;  the  side  CD  =  21.25  chains,  and  the  angle  D  91°  20^ ;  and 
the  side  DE  =  27.41  chains.     Required  the  area  of  the  field. 

Ans.  Go  A.  2  R.  24  P. 

Proh.  20.  From  a  station,  H,  near  the  middle  of  a  field, 
ABCDE F,  from  which  I  could  see  all  the  angles,  I  measured 
the  distances  to  the  several  corners,  and  measured  the  angles 
formed  at  H  by  those  distances,  as  follows : 


Distances. 

Angles. 

AH,  43.15  chains; 

AHB,  60°  30^ 

BH,  29.82       " 

BHC,  47    40 

CH,  35.61       " 

CHD,49    50 

DH,  50.10       " 

DHE,57    10 

EH,  46.18       " 

EHF,64    15 

* 

FH,  36.06       " 

FHA,  80    35 

ired  the  area  of  the  field. 

Ans.  412  A. 

1R.17P. 

NAVIGATION. 
Frob.  1.  From  a  ship  at  sea  I  observed  a  point  of  land  to 
bear  east  by  south,  and,  after  sailing  northeast  12  miles,  I  ob- 


Examples   for   Practice.  189 

served  again,  and  found  its  bearing  to  be  southeast  by  east. 
How  far  was  the  last  observation  made  from  the  point  of  land  ? 

Ans.  26.07  miles. 

I^rob.  2.  If  a  ship  in  latitude  50°  N.,  sails  52  miles  in  the  di- 
rection southwest  by  south,  what  latitude  has  she  arrived  in, 
and  how  much  farther  to  the  west  ? 

Ans.  Latitude  49°  16.^8  N. ;  west,  28.9  miles. 

Prob.  3..  Two  ships  sail  from  the  same  port ;  the  one  sails 
east-northeast  85  miles,  the  other  sails  east  by  south  till  the  first 
ship  bears  northwest  by  west.  What  is  the  distance  of  the  sec- 
ond ship  from  the  port,  and  also  from  the  first  ship  ? 

Ans.  From  the  port,  184.7  miles;   from  the  first  ship, 
123.4  miles. 

Prob.  4.  Two  ports  lie  east  and  west  of  each  other ;  a  ship 
sails  from  each,  namely,  the  ship  from  the  west  port  sails  north- 
east 89  leagues,  and  the  other  sails  80  leagues,  when  she  meets 
the  former.  Required  the  latter  ship's  course,  and  the  distance 
between  the  two  ports. 

Ans.  Course,  N.  51°  52^  W. ;  distance,  112.3  leagues. 

JProb.  5.  Two  ships  sail  from  a  certain  port ;  the  one  sails 
south  by  east  45  leagues,  and  the  other  south-southwest  64 
leagues.  What  is  the  bearing  and  distance  of  the  first  ship  from 
the  second  ? 

Ans.  Bearing,  N.  65°  44^  E. ;  distance,  36.5  leagues. 

Ibvb.  6.  A  ship  sailing  northwest,  two  islands  appear  in 
sight,  of  which  the  one  bears  north,  and  the  other  west-north- 
west; but,  af:er  sailing  20  leagues,  the  former  bears  northeast, 
and  the  latter  west  by  south.  What  is  the  distance  asunder  of 
the  two  islands  ?  Ans.  32.38  leagues. 

Prob.  7.  To  a  vessel  sailing  on  a  certain  course,  a  headland 
was  observed  to  bear  due  west ;  four  hours  after  which  it  was 
seen  at  west-southwest ;  and  six  hours  after  this,  the  vessel  con- 
tinuing to  run  at  the  same  rate,  its  bearing  was  found  to  be 
south-southwest.     What  was  the  vessel's  course  at  the  time  ? 

Ans.  N.  42°  35"  W. 

Prob.  8.  Two  ships  of  war,  intending  to  cannonade  a  fort, 
are,  by  the  shallowness  of  the  water,  kept  so  far  from  it  that  they 
suspect  their  guns  can  not  reach  it  with  effect.  In  order,  there- 
fore, to  measure  the  distance,  they  separate  from  each  other  500 


190  Trigonometry. 

rods;  then  each  ship  observes  the  angle  which  the  other  ship 
and  the  fort  subtend,  which  angles  are  38°  16^  and  37°  9\ 
What,  then,  is  the  distance  between  each  ship  and  the  fort  ? 

Ans.  312  rods  and  320  rods. 
Prob.  9.  A  ship  from  the  latitude  42°  18'  N.,  sails  southwest 
by  south  until  her  latitude  is  40°  18'  N.     What  direct  distance 
has  she  sailed,  and  how  many  miles  has  she  sailed  to  the  west- 
ward ? 

Ans.  Distance  run  144.3  miles,  and  has  sailed  to  west- 
ward 80.2  miles. 
I^roh.  10.  A  ship  having  run  due  east  for  three  days,  at  the 
rate  of  eight  knots  an  hour,  finds  she  has  altered  her  longitude 
15  degrees.     What  parallel  of  latitude  did  she  sail  on  ? 

Ans.  Latitude  50°  12'. 
JProb.  11.  A  ship  in  latitude  43°  30'  N.,  and  longitude  44^ 
W.,  sails  southeasterly  532  miles,  until  her  departure  from  the 
meridian  is  420  miles.     Required  the  course  steered,  and  the 
latitude  and  longitude  of  the  ship. 

Ans.  Course  S.  52°  8'  E.,  latitude  38°  3.'5  N., 
longitude  34°  45'  W. 
Prob.  12.  A  ship  from  latitude  43°  20'  N.,  and  longitude  5?° 
W.,  sails  E.S.E.  until  her  departure  is  745  miles.     Required 
the  distance  sailed,  and  the  latitude  and  longitude  of  the  ship. 
Ans.  Distance  806.4  miles,  latitude  38°  11.'5  N., 
longitude  35°  36'  W. 
Prob.  13.  If  the  height  of  the  mountain  called  the  Peak  of 
Teneriffe  be  4  miles,  and  the  angle  taken  at  the  top  of  it,  as 
formed  between  a  plumb-line  and  a  line  conceived  to  touch  the 
earth  in  the  horizon,  or  farthest  visible  point,  be  87°  25'  55^%  it 
is  required  from  hence  to  determine  the  magnitude  of  the  whole 
earth,  and  the  utmost  distance  that  can  be  seen  on  its  surface 
from  the  top  of  the  mountain,  supposing  the  earth  to  be  a  per- 
fect sphere. 

Ans.  Distance  178.458  miles,  diameter  7957.793  miles. 
Prob.  14.  Required  the  course  and  distance  from  St.  Jago, 
one  of  the  Cape  Verd  islands,  in  latitude  14°  56'  N.,  to  the  island 
of  St.  Helena,  in  latitude  15°  45'  S.,  their  difference  of  longitude 
being  30°  12'. 

Ans.  Course  S.  44°  12'  E.,  distance  2567.8  miles. 


Examples   for   Practice.  191 

Proh.  15.  A  ship  from  the  latitude  of  49°  51'  N.,  and  longi- 
tude of  30°  W.,  sails  S.  39°  W.,  till  she  arrives  in  the  latitude 
of  45°  31^  N.  Required  the  distance  run,  and  the  longitude  of 
the  ship. 

Ans.  Distance  342.3  miles,  longitude  of  ship  35°  2V  W. 
jProb,  16.  Find  the  bearing  and  distance  from  San  Francisco, 
latitude  37°  48^  N.,  longitude  122°  28^  W.,  to  Jeddo,  latitude 
35°  40^  N.,  longitude  139°  40^  E.,  by  Mercator's  sailing. 

Ans.  Course  S.  88°  26^  W.,  distance  4705  miles. 
Proh,  17.  Find  the  bearing  and  distance  from  San  Francisco 
to  Batavia  in  Java,  latitude  6°  9^  S.,  longitude  106°  53^  E.,  by 
Mercator's  sailing. 

Ans.  Course  S.  70°  12^  W.,  distance  7783  miles. 
Prob.  18.  Find  the  bearing  and  distance  from  San  Francisco 
to  Port  Jackson,  latitude  33°  51^  S.,  longitude  151°  14^  E.,  by 
Mercator's  sailing. 

Ans.  Course  S.  48°  18^  W.,  distance  6462  miles. 
Proh.  19.  Find  the  bearing  and  distance  from  San  Francisco 
to  Otaheite,  latitude  17°  29^  S.,  longitude  149°  29^W.,  by  Mer- 
cator's sailing. 

Ans.  Course  S.  24°  44^  W.,  distance  3652  miles. 
Proh.  20.  Find  the  bearing  and  distance  from  San  Francisco 
to  Valparaiso,  latitude  33°  2'  S.,  longitude  71°  41^  W.,  by  Mer- 
cator's sailing. 

Ans.  Course  S.  33°  47^  E.,  distance  5354  miles. 

SPHERICAL  TRIGONOMETRY. 

Proh.  1.  In  the  right-angled  spherical  triangle  ABC,  there  are 
given  the  angle  C  23°  27^  42^^  and  the  side  b  10°  39^  40'^ 
Required  the  angle  B,  and  the  sides  a  and  c. 

r  t»  =11°  35M9^^ 
Ans.  )  c  =  4°  35^  26"^ 
(B=66°58^    V\ 
Proh.  2.  In  the  spherical  triangle  ABC,  the  side  BC  =  90°, 
the  side  AB  =  32°  57^  6^^  and  the  side  AC  =  m^'  32'.    Required 
the  angles. 

^A=132°    2^44^^ 

Ans.  ]  B=  42°  56'  12^^ 

(C=  23°49^26''^ 


192  Trigonometry. 

Pr'oh.  3.  In  the  right-angled  spherical  triangle  ABC,  there 
are  given  the  angle  B  =  47°  54^  20^^  and  the  angle  C^ei^  50^ 
29^^.     Required  the  sides. 

^^  =  61°    4^56^^ 
Arts,  ]6=40o  30^  20'^ 
(  c  r=50o  30^  30^^ 
Proh,  4.  In  the  spherical  triangle  ABC,  the  side  AC  =  90°, 
the  side  AB=:115o  9^  and  the  angle  B^lOi^  40^     Required 
the  remaining  parts. 

(  BC  =  113o  18^    T\ 
Ans,  ]a    =:115°54M6^^ 
(C    =117o33M9'^ 
Proh.  5.  In  the  spherical  triangle  ABC,  the  angle  A  =  130° 
5'  22^^  the  angle  C  =  36°  45^  28^^  and  the  side  AC  =44°  13^ 
45^^-     Required  the  remaining  parts. 

CAB  =  51°    Q'12'\ 
Ans.  ]bC  =  84°  14^  29^^ 
(     B  =  32°26^    6^". 
Prob.  6.  In  the  spherical  triangle  ABC,  the  angle  A=33°  15'' 
7^  B=31°  34^  38^^  and  C  =  161°  25^  17^^    Required  the  sides. 

(a=  80°    5^    4^^ 
Ans.  \b=  70°  10^  30^^ 
(c=145°    5^    2^ 
Prob.  7.  In  the  spherical  triangle  ABC,  the  side  AB  =  112° 
22^  58^  AC  =  52°  39^  4^^  and  BC  =  89°  m  53''.     Required 
the  angles. 

C  \=  70°  39^  0'\ 
Ans.  <  B  =  48°  36'  0''. 
(  C=119°  15' 0'^ 
Prob.  8.  In  the  spherical  triangle  ABC,  the  side  AB  =  76°  35' 
36",  AC  =  50°  10'  30",  and  the  angle  A =34°  15'  3".     Re- 
quired the  remaining  parts. 

(  B    =42°  15'  13". 
Ans.  ]c    =121°  36' 20". 
(  BC=  40°    0'  10". 
Prob.  9.  The  latitudes  of  the  observatories  of  Paris  and  Pekin 
are  48°  50'  14"  N.  and  39°  54'  13"  N.,  and  their  difference  of 
longitude  is  114°  7'  30".     What  is  their  distance  ? 

Ans.  73°  56'  40". 


Examples   for   Practice.  193 

Proh.  10.  Required  the  course  and  distance  from  New  York, 
latitude  40^  43"  N.,  longitude  74°  0"  W.,  to  San  Francisco,  lat- 
itude 370  48"  N.,  longitude  122°  28"  W.,  on  the  shortest  route. 
Ans.  The  course  is  S.  78°  16"  W. 

Distance,  2229.8  nautical  miles. 
Proh.  11.  Required  the  course  and  distance  from  San  Fran- 
cisco, latitude  37°  48"  N.,  longitude  122°  28"  W.,  to  Jeddo,  in 
latitude  35°  40"  N.,  longitude  139°  40"  E.,  on  the  shortest 
route.  Ans.  The  course  is  N.  b&^  41"  W. 

Distance,  4461.9  nautical  miles. 
Prob.  12.  Required  the  course  and  distance  from  San  Fran- 
cisco to  Batavia  in  Java,  latitude  6°  9"  S.,  longitude  106°  53"  E., 
on  the  shortest  route. 

Ans.  The  course  is  N.  67°  30"  W. 
Distance,  7516  nautical  miles. 
P^ob.  13.  Required  the  course  and  distance  from  San  Fran- 
cisco to  Port  Jackson,  latitude  33°  51"  S.,  longitude  151°  14"  E., 
on  the  shortest  route. 

Ans.  The  course  is  S.  59°  50"  W. 
Distance,  6444  nautical  miles. 
Prob.  14.  Required  the  course  and  distance  from  San  Fran- 
cisco to  Otaheite,  latitude  17°  29"  S.,  longitude  149°  29"  W.,  on 
the  shortest  route. 

Ans.  The  course  is  S.  30°  31"  W. 

Distance,  3650.3  nautical  miles. 
Prob.  15.  Required  the  course  and  distance  from  San  Fran- 
cisco to  Valparaiso,  latitude  33°  2"  S.,  longitude  71°  41  W.,  on 
the  shortest  route. 

Ans.  The  course  is  S.  5^^  9"  E. 

Distance,  5108.5  nautical  miles. 
Prob.  16.  Suppose  two  ports,  one  in  north  latitude  30°,  and 
the  other  in  north  latitude  40°,  the  difference  of  longitude  be- 
tween them  being  50°.     Required  the  bearing  and  distance 
from  each  of  these  ports  to  an  island  that  lies  in  south  latitude 
18°,  and  which  is  equally  distant  from  both  of  the  said  ports. 
Ans.  Bearing  from  first  port,      S.  40°  52"    9""  E. 
Bearing  from  second  port,  S.  15      9  47  W. 
The  distance,  59°  23"  19""  =  3563.3  nautical  miles. 

THE    END. 

N' 


I 


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